Solve 2x^2-2x-5x=2

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

Solve for x

$$x = \frac{\sqrt{65} + 7}{4} \approx 3.765564437$$

$$x=\frac{7-\sqrt{65}}{4}\approx -0.265564437$$

Steps Using the Quadratic Formula

Combine -2x and -5x to get -7x.

$$2x^{2}-7x=2$$

Subtract 2 from both sides.

$$2x^{2}-7x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -7 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\left(-2\right)}}{2\times 2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 2\left(-2\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-8\left(-2\right)}}{2\times 2}$$

Multiply -8 times -2.

$$x=\frac{-\left(-7\right)±\sqrt{49+16}}{2\times 2}$$

Add 49 to 16.

$$x=\frac{-\left(-7\right)±\sqrt{65}}{2\times 2}$$

The opposite of -7 is 7.

$$x=\frac{7±\sqrt{65}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{7±\sqrt{65}}{4}$$

Now solve the equation x=\frac{7±\sqrt{65}}{4} when ± is plus. Add 7 to \sqrt{65}.

$$x=\frac{\sqrt{65}+7}{4}$$

Now solve the equation x=\frac{7±\sqrt{65}}{4} when ± is minus. Subtract \sqrt{65} from 7.

$$x=\frac{7-\sqrt{65}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{65}+7}{4} x=\frac{7-\sqrt{65}}{4}$$

Steps for Completing the Square

Combine -2x and -5x to get -7x.

$$2x^{2}-7x=2$$

Divide both sides by 2.

$$\frac{2x^{2}-7x}{2}=\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{7}{2}x=\frac{2}{2}$$

Divide 2 by 2.

$$x^{2}-\frac{7}{2}x=1$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=1+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=1+\frac{49}{16}$$

Add 1 to \frac{49}{16}.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{65}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{65}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{65}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{\sqrt{65}}{4} x-\frac{7}{4}=-\frac{\sqrt{65}}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=\frac{\sqrt{65}+7}{4} x=\frac{7-\sqrt{65}}{4}$$