Solve Using the Quadratic Formula 6x^2+11x-35=0

$6 x^{2} + 11 x - 35 = 0$

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 $a = 6$ , $b = 11$ , and $c = - 35$ into the quadratic formula and solve for $x$ .

$\frac{- 11 \pm \sqrt{11^{2} - 4 \cdot (6 \cdot - 35)}}{2 \cdot 6}$

Step 3

Simplify.

Simplify the numerator.

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

$x = \frac{- 11 \pm \sqrt{121 - 4 \cdot 6 \cdot - 35}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot - 35$ .

Multiply $- 4$ by $6$ .

$x = \frac{- 11 \pm \sqrt{121 - 24 \cdot - 35}}{2 \cdot 6}$

Multiply $- 24$ by $- 35$ .

$x = \frac{- 11 \pm \sqrt{121 + 840}}{2 \cdot 6}$

Add $121$ and $840$ .

$x = \frac{- 11 \pm \sqrt{961}}{2 \cdot 6}$

Rewrite $961$ as $31^{2}$ .

$x = \frac{- 11 \pm \sqrt{31^{2}}}{2 \cdot 6}$

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

$x = \frac{- 11 \pm 31}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = \frac{- 11 \pm 31}{12}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{5}{3}, - \frac{7}{2}$