We will use the Quadratic Formula to solve the given quadratic equation.

a x^{2} + b x + c = 0 \Leftrightarrow x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

We first need to identify the values of

a,

b,

and

c .

x^{2} + 12 x - 9 = 0 \Leftrightarrow 1 x^{2} + 12 x + (- 9) = 0

We see that

a = 1,

b = 12,

and

c = - 9 .

Let's substitute these values into the Quadratic Formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Substitute values

x = \frac{- 1 2 \pm 1 2 ^{2} - 4 ( 1 ) ( - 9 )}{2 ( 1 )}

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Solve for

x

and Simplify

Calculate power

x = \frac{- 1 2 \pm 1 4 4 - 4 ( 1 ) ( - 9 )}{2 ( 1 )}

$a \cdot 1 = a$

x = \frac{- 1 2 \pm 1 4 4 - 4 ( - 9 )}{2}

$- a (- b) = a \cdot b$

x = \frac{- 1 2 \pm 1 4 4 + 3 6}{2}

Add terms

x = \frac{- 1 2 \pm 1 8 0}{2}

Split into factors

x = \frac{- 1 2 \pm 3 6 \cdot 5}{2}

$a \cdot b = a \cdot b$

x = \frac{- 1 2 \pm 3 6 \cdot 5}{2}

Calculate root

x = \frac{- 1 2 \pm 6 5}{2}

Factor out $2$

x = \frac{2 ( - 6 \pm 3 5 )}{2}

Cancel out common factors

x = - 6 \pm 35

Using the Quadratic Formula, we found that the solutions of the given equation are

x = - 6 \pm 35 .

Therefore, the solutions are

x_{1} = - 6 + 35

and

x_{2} = - 6 - 35 .

Exercise