Solve Using the Quadratic Formula 9a=-a^2-18

$9 a = - a^{2} - 18$

Step 1

Move all the expressions to the left side of the equation.

Add $a^{2}$ to both sides of the equation.

$9 a + a^{2} = - 18$

Add $18$ to both sides of the equation.

$9 a + a^{2} + 18 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = 9$ , and $c = 18$ into the quadratic formula and solve for $a$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (1 \cdot 18)}}{2 \cdot 1}$

Step 4

Simplify.

Simplify the numerator.

Raise $9$ to the power of $2$ .

$a = \frac{- 9 \pm \sqrt{81 - 4 \cdot 1 \cdot 18}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 18$ .

Multiply $- 4$ by $1$ .

$a = \frac{- 9 \pm \sqrt{81 - 4 \cdot 18}}{2 \cdot 1}$

Multiply $- 4$ by $18$ .

$a = \frac{- 9 \pm \sqrt{81 - 72}}{2 \cdot 1}$

Subtract $72$ from $81$ .

$a = \frac{- 9 \pm \sqrt{9}}{2 \cdot 1}$

Rewrite $9$ as $3^{2}$ .

$a = \frac{- 9 \pm \sqrt{3^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$a = \frac{- 9 \pm 3}{2 \cdot 1}$

Multiply $2$ by $1$ .

$a = \frac{- 9 \pm 3}{2}$

Step 5

The final answer is the combination of both solutions.

$a = - 3, - 6$