Solve 2x^2=x+5

$$2 x ^ { 2 } = x + 5$$

Solve for x

$$x = \frac{\sqrt{41} + 1}{4} \approx 1.850781059$$

$$x=\frac{1-\sqrt{41}}{4}\approx -1.350781059$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$2x^{2}-x=5$$

Subtract 5 from both sides.

$$2x^{2}-x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-5\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\left(-5\right)}}{2\times 2}$$

Multiply -8 times -5.

$$x=\frac{-\left(-1\right)±\sqrt{1+40}}{2\times 2}$$

Add 1 to 40.

$$x=\frac{-\left(-1\right)±\sqrt{41}}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±\sqrt{41}}{4}$$

Now solve the equation x=\frac{1±\sqrt{41}}{4} when ± is plus. Add 1 to \sqrt{41}.

$$x=\frac{\sqrt{41}+1}{4}$$

Now solve the equation x=\frac{1±\sqrt{41}}{4} when ± is minus. Subtract \sqrt{41} from 1.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract x from both sides.

$$2x^{2}-x=5$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{1}{2}x=\frac{5}{2}$$

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=\frac{5}{2}+\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{5}{2}+\frac{1}{16}$$

Add \frac{5}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{41}{16}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{41}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Simplify.

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

Add \frac{1}{4} to both sides of the equation.

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