Find the Roots (Zeros) -12x^2+11x-3=0

$- 12 x^{2} + 11 x - 3 = 0$

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 $a = - 12$ , $b = 11$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{- 11 \pm \sqrt{11^{2} - 4 \cdot (- 12 \cdot - 3)}}{2 \cdot - 12}$

Step 3

Simplify.

Simplify the numerator.

Raise $11$ to the power of $2$ .

$x = \frac{- 11 \pm \sqrt{121 - 4 \cdot - 12 \cdot - 3}}{2 \cdot - 12}$

Multiply $- 4 \cdot - 12 \cdot - 3$ .

Multiply $- 4$ by $- 12$ .

$x = \frac{- 11 \pm \sqrt{121 + 48 \cdot - 3}}{2 \cdot - 12}$

Multiply $48$ by $- 3$ .

$x = \frac{- 11 \pm \sqrt{121 - 144}}{2 \cdot - 12}$

Subtract $144$ from $121$ .

$x = \frac{- 11 \pm \sqrt{- 23}}{2 \cdot - 12}$

Rewrite $- 23$ as $- 1 (23)$ .

$x = \frac{- 11 \pm \sqrt{- 1 \cdot 23}}{2 \cdot - 12}$

Rewrite $\sqrt{- 1 (23)}$ as $\sqrt{- 1} \cdot \sqrt{23}$ .

$x = \frac{- 11 \pm \sqrt{- 1} \cdot \sqrt{23}}{2 \cdot - 12}$

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

$x = \frac{- 11 \pm i \sqrt{23}}{2 \cdot - 12}$

Multiply $2$ by $- 12$ .

$x = \frac{- 11 \pm i \sqrt{23}}{- 24}$

Simplify $\frac{- 11 \pm i \sqrt{23}}{- 24}$ .

$x = \frac{11 \pm i \sqrt{23}}{24}$

Step 4