Find All Complex Solutions 9a^2+25a+16=a

$9 a^{2} + 25 a + 16 = a$

Step 1

Move all terms to the left side of the equation and simplify.

Subtract $a$ from both sides of the equation.

$9 a^{2} + 25 a + 16 - a = 0$

Subtract $a$ from $25 a$ .

$9 a^{2} + 24 a + 16 = 0$

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 $a = 9$ , $b = 24$ , and $c = 16$ into the quadratic formula and solve for $a$ .

$\frac{- 24 \pm \sqrt{24^{2} - 4 \cdot (9 \cdot 16)}}{2 \cdot 9}$

Step 4

Simplify.

Simplify the numerator.

Raise $24$ to the power of $2$ .

$a = \frac{- 24 \pm \sqrt{576 - 4 \cdot 9 \cdot 16}}{2 \cdot 9}$

Multiply $- 4 \cdot 9 \cdot 16$ .

Multiply $- 4$ by $9$ .

$a = \frac{- 24 \pm \sqrt{576 - 36 \cdot 16}}{2 \cdot 9}$

Multiply $- 36$ by $16$ .

$a = \frac{- 24 \pm \sqrt{576 - 576}}{2 \cdot 9}$

Subtract $576$ from $576$ .

$a = \frac{- 24 \pm \sqrt{0}}{2 \cdot 9}$

Rewrite $0$ as $0^{2}$ .

$a = \frac{- 24 \pm \sqrt{0^{2}}}{2 \cdot 9}$

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

$a = \frac{- 24 \pm 0}{2 \cdot 9}$

$- 24$ plus or minus $0$ is $- 24$ .

$a = \frac{- 24}{2 \cdot 9}$

Multiply $2$ by $9$ .

$a = \frac{- 24}{18}$

Cancel the common factor of $- 24$ and $18$ .

Factor $6$ out of $- 24$ .

$a = \frac{6 (- 4)}{18}$

Cancel the common factors.

Factor $6$ out of $18$ .

$a = \frac{6 \cdot - 4}{6 \cdot 3}$

Cancel the common factor.

$a = \frac{6 \cdot - 4}{6 \cdot 3}$

Rewrite the expression.

$a = \frac{- 4}{3}$

Move the negative in front of the fraction.

$a = - \frac{4}{3}$

Step 5

The final answer is the combination of both solutions.

$a = - \frac{4}{3}$ Double roots

Find All Complex Solutions 9a^2+25a+16=a

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