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

$6 x^{2} + x = 15$

Step 1

Subtract $15$ from both sides of the equation.

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

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 6$ , $b = 1$ , and $c = - 15$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (6 \cdot - 15)}}{2 \cdot 6}$

Step 4

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 6 \cdot - 15}}{2 \cdot 6}$

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

Multiply $- 4$ by $6$ .

$x = \frac{- 1 \pm \sqrt{1 - 24 \cdot - 15}}{2 \cdot 6}$

Multiply $- 24$ by $- 15$ .

$x = \frac{- 1 \pm \sqrt{1 + 360}}{2 \cdot 6}$

Add $1$ and $360$ .

$x = \frac{- 1 \pm \sqrt{361}}{2 \cdot 6}$

Rewrite $361$ as $19^{2}$ .

$x = \frac{- 1 \pm \sqrt{19^{2}}}{2 \cdot 6}$

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

$x = \frac{- 1 \pm 19}{2 \cdot 6}$

Multiply $2$ by $6$ .

$x = \frac{- 1 \pm 19}{12}$

Step 5

The final answer is the combination of both solutions.

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