Solve 2x^2=-10x-7

$$2 x ^ { 2 } = - 10 x - 7$$

Solve for x

$$x=\frac{\sqrt{11}-5}{2}\approx -0.841687605$$

$$x=\frac{-\sqrt{11}-5}{2}\approx -4.158312395$$

Steps Using the Quadratic Formula

Add 10x to both sides.

$$2x^{2}+10x=-7$$

Add 7 to both sides.

$$2x^{2}+10x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 10 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 2\times 7}}{2\times 2}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 2\times 7}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-10±\sqrt{100-8\times 7}}{2\times 2}$$

Multiply -8 times 7.

$$x=\frac{-10±\sqrt{100-56}}{2\times 2}$$

Add 100 to -56.

$$x=\frac{-10±\sqrt{44}}{2\times 2}$$

Take the square root of 44.

$$x=\frac{-10±2\sqrt{11}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-10±2\sqrt{11}}{4}$$

Now solve the equation x=\frac{-10±2\sqrt{11}}{4} when ± is plus. Add -10 to 2\sqrt{11}.

$$x=\frac{2\sqrt{11}-10}{4}$$

Divide -10+2\sqrt{11} by 4.

$$x=\frac{\sqrt{11}-5}{2}$$

Now solve the equation x=\frac{-10±2\sqrt{11}}{4} when ± is minus. Subtract 2\sqrt{11} from -10.

$$x=\frac{-2\sqrt{11}-10}{4}$$

Divide -10-2\sqrt{11} by 4.

$$x=\frac{-\sqrt{11}-5}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{11}-5}{2} x=\frac{-\sqrt{11}-5}{2}$$

Steps for Completing the Square

Add 10x to both sides.

$$2x^{2}+10x=-7$$

Divide both sides by 2.

$$\frac{2x^{2}+10x}{2}=-\frac{7}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{10}{2}x=-\frac{7}{2}$$

Divide 10 by 2.

$$x^{2}+5x=-\frac{7}{2}$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-\frac{7}{2}+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=-\frac{7}{2}+\frac{25}{4}$$

Add -\frac{7}{2} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+5x+\frac{25}{4}=\frac{11}{4}$$

Factor x^{2}+5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{11}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{11}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{\sqrt{11}}{2} x+\frac{5}{2}=-\frac{\sqrt{11}}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=\frac{\sqrt{11}-5}{2} x=\frac{-\sqrt{11}-5}{2}$$