Solve Using the Quadratic Formula 2x^2+x=10

$2 x^{2} + x = 10$

Step 1

Subtract $10$ from both sides of the equation.

$2 x^{2} + x - 10 = 0$

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 $a = 2$ , $b = 1$ , and $c = - 10$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (2 \cdot - 10)}}{2 \cdot 2}$

Step 4

Simplify.

Simplify the numerator.

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 2 \cdot - 10}}{2 \cdot 2}$

Multiply $- 4 \cdot 2 \cdot - 10$ .

Multiply $- 4$ by $2$ .

$x = \frac{- 1 \pm \sqrt{1 - 8 \cdot - 10}}{2 \cdot 2}$

Multiply $- 8$ by $- 10$ .

$x = \frac{- 1 \pm \sqrt{1 + 80}}{2 \cdot 2}$

Add $1$ and $80$ .

$x = \frac{- 1 \pm \sqrt{81}}{2 \cdot 2}$

Rewrite $81$ as $9^{2}$ .

$x = \frac{- 1 \pm \sqrt{9^{2}}}{2 \cdot 2}$

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

$x = \frac{- 1 \pm 9}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{- 1 \pm 9}{4}$

Step 5

The final answer is the combination of both solutions.

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