Solve Using the Quadratic Formula 6x^2+3x-15=0

$6 x^{2} + 3 x - 15 = 0$

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 $a = 6$ , $b = 3$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (6 \cdot - 15)}}{2 \cdot 6}$

Step 3

Simplify.

Simplify the numerator.

Raise $3$ to the power of $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 6 \cdot - 15}}{2 \cdot 6}$

Multiply $- 4 \cdot 6 \cdot - 15$ .

Multiply $- 4$ by $6$ .

$x = \frac{- 3 \pm \sqrt{9 - 24 \cdot - 15}}{2 \cdot 6}$

Multiply $- 24$ by $- 15$ .

$x = \frac{- 3 \pm \sqrt{9 + 360}}{2 \cdot 6}$

Add $9$ and $360$ .

$x = \frac{- 3 \pm \sqrt{369}}{2 \cdot 6}$

Rewrite $369$ as $3^{2} \cdot 41$ .

Factor $9$ out of $369$ .

$x = \frac{- 3 \pm \sqrt{9 (41)}}{2 \cdot 6}$

Rewrite $9$ as $3^{2}$ .

$x = \frac{- 3 \pm \sqrt{3^{2} \cdot 41}}{2 \cdot 6}$

Pull terms out from under the radical.

$x = \frac{- 3 \pm 3 \sqrt{41}}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = \frac{- 3 \pm 3 \sqrt{41}}{12}$

Simplify $\frac{- 3 \pm 3 \sqrt{41}}{12}$ .

$x = \frac{- 1 \pm \sqrt{41}}{4}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{- 1 \pm \sqrt{41}}{4}$

Decimal Form:

$x = 1.35078105 \dots, - 1.85078105 \dots$