Solve Using the Quadratic Formula 6x^2-7x-3=0

$6 x^{2} - 7 x - 3 = 0$

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 $a = 6$ , $b = - 7$ , and $c = - 3$ into the quadratic formula and solve for $x$ .

$\frac{7 \pm \sqrt{{(- 7)}^{2} - 4 \cdot (6 \cdot - 3)}}{2 \cdot 6}$

Step 3

Simplify.

Simplify the numerator.

Raise $- 7$ to the power of $2$ .

$x = \frac{7 \pm \sqrt{49 - 4 \cdot 6 \cdot - 3}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot - 3$ .

Multiply $- 4$ by $6$ .

$x = \frac{7 \pm \sqrt{49 - 24 \cdot - 3}}{2 \cdot 6}$

Multiply $- 24$ by $- 3$ .

$x = \frac{7 \pm \sqrt{49 + 72}}{2 \cdot 6}$

Add $49$ and $72$ .

$x = \frac{7 \pm \sqrt{121}}{2 \cdot 6}$

Rewrite $121$ as $11^{2}$ .

$x = \frac{7 \pm \sqrt{11^{2}}}{2 \cdot 6}$

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

$x = \frac{7 \pm 11}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = \frac{7 \pm 11}{12}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{3}{2}, - \frac{1}{3}$