Solve for x 2x^2=x+9

$2 x^{2} = x + 9$

Subtract $x$ from both sides of the equation.

$2 x^{2} - x = 9$

Subtract $9$ from both sides of the equation.

$2 x^{2} - x - 9 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 2$ , $b = - 1$ , and $c = - 9$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (2 \cdot - 9)}}{2 \cdot 2}$

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 2 \cdot - 9}}{2 \cdot 2}$

Multiply $- 4$ by $2$ .

$x = \frac{1 \pm \sqrt{1 - 8 \cdot - 9}}{2 \cdot 2}$

Multiply $- 8$ by $- 9$ .

$x = \frac{1 \pm \sqrt{1 + 72}}{2 \cdot 2}$

Add $1$ and $72$ .

$x = \frac{1 \pm \sqrt{73}}{2 \cdot 2}$

Multiply $2$ by $2$ .

$x = \frac{1 \pm \sqrt{73}}{4}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{73}}{4}, \frac{1 - \sqrt{73}}{4}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 + \sqrt{73}}{4}, \frac{1 - \sqrt{73}}{4}$

Decimal Form:

$x = 2.38600093 \dots, - 1.88600093 \dots$