Another method for solving a quadratic equation is to use the quadratic formula. To use the formula, the equation must be in the standard form of a quadratic equation. The formula is shown below.

x = fraction; numerator: -b +- square root of (b squared minus 4ac), denominator: 2a

Example:

x2 + 6x = 4

x2 + 6x – 4 = 0

first equation: x = fraction; numerator: -6 +- square root of (6 squared - 4(1)(-4)), denominator: 2(1) second equation: x equations fraction; numerator: -6 +- square root of (36 + 16), denominator: 2 third equation: x = fraction; numerator: -6 +- square root of 52, denominator: 2

x ≈ -6 ± 7.21 2

x ≈ -6 + 7.21 2  or x ≈ -6 – 7.21 2

x ≈ 0.605            x ≈ -6.605

The first step is to put the quadratic equation in standard form.

Identify the values of a, b, and c and use these values in the formula. a = 1, b = 6, c = -4

Begin simplifying by performing the operations under the square root.


Finish simplifying the expression.


Approximate √52 to the nearest hundredths.

The solutions to the equation are approximately 0.605 and -0.605.

Example:

3x2 – 4x – 2 = 0

The equation is already in standard form. Identify a, b, and c: Use these values in the quadratic formula.

first line: x = fraction; numerator: -(-4) +- square root of ((-4) squared - 4(3)(-2), denominator: 2(3) second line: x = fraction; numerator: 4 +- square root of (16 + 24), denominator: 6 third line: x = fraction; numerator: 4 +- square root of 40, denominator: 6 fourth line: x equals approximately fraction; numerator: 4 + 6.32, denominator: 6 or x equals approximately fraction; numerator: 4 - 6.32, denominator: 6 fifth line: x equals approximately 1.72 or x equals approximately -0.387

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