Find the Roots (Zeros) -10x^2+12x-9=0

$- 10 x^{2} + 12 x - 9 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = - 10$ , $b = 12$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{- 12 \pm \sqrt{12^{2} - 4 \cdot (- 10 \cdot - 9)}}{2 \cdot - 10}$

Step 3

Simplify.

Simplify the numerator.

Raise $12$ to the power of $2$ .

$x = \frac{- 12 \pm \sqrt{144 - 4 \cdot - 10 \cdot - 9}}{2 \cdot - 10}$

Multiply $- 4 \cdot - 10 \cdot - 9$ .

Multiply $- 4$ by $- 10$ .

$x = \frac{- 12 \pm \sqrt{144 + 40 \cdot - 9}}{2 \cdot - 10}$

Multiply $40$ by $- 9$ .

$x = \frac{- 12 \pm \sqrt{144 - 360}}{2 \cdot - 10}$

Subtract $360$ from $144$ .

$x = \frac{- 12 \pm \sqrt{- 216}}{2 \cdot - 10}$

Rewrite $- 216$ as $- 1 (216)$ .

$x = \frac{- 12 \pm \sqrt{- 1 \cdot 216}}{2 \cdot - 10}$

Rewrite $\sqrt{- 1 (216)}$ as $\sqrt{- 1} \cdot \sqrt{216}$ .

$x = \frac{- 12 \pm \sqrt{- 1} \cdot \sqrt{216}}{2 \cdot - 10}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 12 \pm i \cdot \sqrt{216}}{2 \cdot - 10}$

Rewrite $216$ as $6^{2} \cdot 6$ .

Factor $36$ out of $216$ .

$x = \frac{- 12 \pm i \cdot \sqrt{36 (6)}}{2 \cdot - 10}$

Rewrite $36$ as $6^{2}$ .

$x = \frac{- 12 \pm i \cdot \sqrt{6^{2} \cdot 6}}{2 \cdot - 10}$

Pull terms out from under the radical.

$x = \frac{- 12 \pm i \cdot (6 \sqrt{6})}{2 \cdot - 10}$

Move $6$ to the left of $i$ .

$x = \frac{- 12 \pm 6 i \sqrt{6}}{2 \cdot - 10}$

Multiply $2$ by $- 10$ .

$x = \frac{- 12 \pm 6 i \sqrt{6}}{- 20}$

Simplify $\frac{- 12 \pm 6 i \sqrt{6}}{- 20}$ .

$x = \frac{6 \pm 3 i \sqrt{6}}{10}$

Step 4