Quadratic Formula: Derivation, Discriminant, Examples

The quadratic formula solves ( ax^2 + bx + c = 0 ): ( x = frac{-b pm sqrt{b^2 – 4ac}}{2a} ). The discriminant ( b^2 – 4ac ) gives the number of real solutions: two if positive, one if zero, none (two complex) if negative. Example: ( 2x^2 – 7x + 3 = 0 ) gives ( x = 3 ) or ( x = frac12 ).

The quadratic formula solves any quadratic equation ( ax^2 + bx + c = 0 ) with ( a neq 0 ):

$$x = frac{-b pm sqrt{b^2 - 4ac}}{2a}$$

The ( pm ) gives two solutions: one with plus, one with minus. Unlike factoring, the formula always works, whether the roots are whole numbers, irrational, or complex.

How to use the quadratic formula

1. Write the equation in standard form, ( ax^2 + bx + c = 0 ), with zero on one side.
2. Identify a, b and c, including their signs.
3. Compute the discriminant ( b^2 – 4ac ).
4. Substitute into the formula and simplify both roots.

Example 1: two rational roots

Solve ( 2x^2 – 7x + 3 = 0 ). Here ( a = 2 ), ( b = -7 ), ( c = 3 ).

$$x = frac{7 pm sqrt{49 - 24}}{4} = frac{7 pm 5}{4}$$

So ( x = frac{12}{4} = 3 ) or ( x = frac{2}{4} = frac12 ).

Example 2: irrational roots

Solve ( x^2 – 4x – 1 = 0 ): ( a = 1 ), ( b = -4 ), ( c = -1 ).

$$x = frac{4 pm sqrt{16 + 4}}{2} = frac{4 pm sqrt{20}}{2} = frac{4 pm 2sqrt5}{2} = 2 pm sqrt5$$

That’s ( x approx 4.236 ) and ( x approx -0.236 ).

Example 3: rearranging first

Solve ( 3x^2 = 5x + 2 ). Move everything to one side: ( 3x^2 – 5x – 2 = 0 ). Then

$$x = frac{5 pm sqrt{25 + 24}}{6} = frac{5 pm 7}{6}$$

so ( x = 2 ) or ( x = -frac13 ).

Example 4: complex roots

Solve ( x^2 + 2x + 5 = 0 ). The discriminant is ( 4 – 20 = -16 ), and ( sqrt{-16} = 4i ):

$$x = frac{-2 pm 4i}{2} = -1 pm 2i$$

The discriminant

The expression under the square root, ( Delta = b^2 – 4ac ), tells you what kind of roots to expect before you finish the calculation.

Discriminant	Number and type of roots	Graph of ( y = ax^2 + bx + c )
Positive perfect square	two rational roots	crosses the x-axis twice
Positive, not a perfect square	two irrational roots	crosses the x-axis twice
Zero	one repeated root, ( x = -frac{b}{2a} )	touches the x-axis at the vertex
Negative	two complex roots	doesn’t meet the x-axis

For example, ( x^2 – 6x + 9 = 0 ) has ( Delta = 36 – 36 = 0 ), so its only root is ( x = 3 ), and indeed it factors as ( (x – 3)^2 ).

Derivation: completing the square

Start with ( ax^2 + bx + c = 0 ) and divide by a:

$$x^2 + frac{b}{a}x = -frac{c}{a}$$

Add ( left( frac{b}{2a} right)^2 ) to both sides so the left side becomes a perfect square:

$$left( x + frac{b}{2a} right)^2 = frac{b^2}{4a^2} - frac{c}{a} = frac{b^2 - 4ac}{4a^2}$$

Take square roots of both sides:

$$x + frac{b}{2a} = pmfrac{sqrt{b^2 - 4ac}}{2a}$$

Subtract ( frac{b}{2a} ), and you have the quadratic formula.

Quadratic formula calculator

Enter a, b and c to see the discriminant, both roots, the factored form and the vertex.

The full-page quadratic formula calculator has more examples.

Quadratic functions and their graphs: the parabola

A quadratic function is ( f(x) = ax^2 + bx + c ) with ( a neq 0 ). Its graph is a parabola.

• It opens up if ( a > 0 ) and down if ( a < 0 ).
• The roots are its x-intercepts, and c is its y-intercept.
• The vertex is at ( x = -frac{b}{2a} ), exactly halfway between the roots. It’s the minimum point when ( a > 0 ) and the maximum when ( a < 0 ).
• The axis of symmetry is the vertical line through the vertex.

In vertex form, ( y = a(x – h)^2 + k ), the vertex ( (h, k) ) can be read off directly.

Example 5: maximum height

A ball’s height is ( h(t) = -5t^2 + 20t + 1 ) metres. The vertex is at ( t = -frac{20}{2(-5)} = 2 ) s, and ( h(2) = -20 + 40 + 1 = 21 ) m, the maximum height. It lands when ( h(t) = 0 ):

$$t = frac{-20 pm sqrt{400 + 20}}{-10} approx 4.05 text{ s}$$

(the other root is negative). This is projectile motion in one dimension.

Other ways to solve quadratics

Method	Best when
Factoring	the roots are small whole numbers or simple fractions
Square roots	there’s no x term: ( x^2 = 9 Rightarrow x = pm3 )
Completing the square	you need vertex form
Quadratic formula	always works; use it when factoring isn’t quick

Quadratics in calculus

Quadratics turn up constantly. Setting the derivative of a cubic to zero gives a quadratic, so finding critical points of ( f(x) = x^3 – 6x^2 + 9x ) means solving ( 3x^2 – 12x + 9 = 0 ), which gives ( x = 1 ) and ( x = 3 ). Quadratics that don’t factor also decide how partial fractions are set up. For higher-degree equations, see polynomials.

Common mistakes

• Sign of b. If ( b = -7 ), then ( -b = 7 ) and ( b^2 = 49 ).
• Partial division. The ( 2a ) divides the whole numerator, ( -b pm sqrt{Delta} ), not just the root.
• Not setting the equation to zero first. You can’t read off c from ( x^2 + 3x = 10 ) until it’s ( x^2 + 3x – 10 = 0 ).
• Simplifying radicals incorrectly. ( frac{4 pm 2sqrt5}{2} = 2 pm sqrt5 ); divide every term by 2.
• Calling negative-discriminant equations unsolvable. They have no real roots, but they do have complex ones.

Practice problems

Solve each equation.

1. ( x^2 – 5x + 6 = 0 )
2. ( x^2 + 4x – 21 = 0 )
3. ( 2x^2 + 3x – 2 = 0 )
4. ( x^2 – 2x – 2 = 0 )
5. ( x^2 + 4x + 13 = 0 )
6. Find the discriminant of ( 4x^2 – 12x + 9 = 0 ) and describe the roots.

Answers: (1) 2, 3; (2) 3, −7; (3) ( frac12 ), −2; (4) ( 1 pm sqrt3 ); (5) ( -2 pm 3i ); (6) 0, one repeated root ( x = frac32 ).

Quadratic equation MCQs

Test yourself; answers are below.

1. The roots of ( x^2 – 7x + 12 = 0 ) are: (a) 3, 4 (b) −3, −4 (c) 2, 6 (d) 1, 12
2. The discriminant of ( 2x^2 + 3x + 5 = 0 ) is: (a) 49 (b) −31 (c) 31 (d) −49
3. ( x^2 + 6x + 9 = 0 ) has: (a) two distinct real roots (b) one repeated root (c) no real roots (d) three roots
4. The sum of the roots of ( 3x^2 – 12x + 5 = 0 ) is: (a) −4 (b) 4 (c) ( frac53 ) (d) 12
5. The vertex of ( y = x^2 – 4x + 1 ) is at: (a) ( (2, -3) ) (b) ( (-2, 13) ) (c) ( (4, 1) ) (d) ( (2, 3) )
6. For which k does ( x^2 + kx + 16 = 0 ) have equal roots? (a) ( pm4 ) (b) ( pm8 ) (c) 16 (d) 0

Answers: (1) a; (2) b, since ( 9 – 40 = -31 ); (3) b, since ( Delta = 0 ); (4) b, since the sum of roots is ( -frac{b}{a} = 4 ); (5) a; (6) b, since ( k^2 – 64 = 0 ).

The sum and product shortcuts in questions 4 and 6 come from Vieta’s formulas: for ( ax^2 + bx + c = 0 ), the roots add to ( -frac{b}{a} ) and multiply to ( frac{c}{a} ).

FAQ

What is the quadratic formula?

( x = frac{-b pm sqrt{b^2 – 4ac}}{2a} ). It gives the solutions of ( ax^2 + bx + c = 0 ).

What is the discriminant?

( b^2 – 4ac ). It tells you whether a quadratic has two real roots, one repeated root, or two complex roots.

When should I use the quadratic formula instead of factoring?

Whenever the equation doesn’t factor quickly. The formula always works; factoring is just faster when it’s easy.

What is a quadratic equation?

An equation in which the highest power of the unknown is 2, such as ( 3x^2 – x + 4 = 0 ).

How do you find the vertex of a quadratic?

Use ( x = -frac{b}{2a} ), then substitute that x into the equation to find y.

Further reading

• Quadratic formula (Wikipedia) — derivations, history and the formula’s geometric meaning.
• Solving Quadratic Equations, Part II (Paul’s Online Math Notes) — completing the square, the quadratic formula and the discriminant.
• Parabolas (Paul’s Online Math Notes) — graphing quadratics, vertex form and intercepts.