Solve 2x^2-8x-4x=11

$$2 x ^ { 2 } - 8 x - 4 x = 11$$

Solve for x

$$x=\frac{\sqrt{58}}{2}+3\approx 6.807886553$$

$$x=-\frac{\sqrt{58}}{2}+3\approx -0.807886553$$

Steps Using the Quadratic Formula

Combine -8x and -4x to get -12x.

$$2x^{2}-12x=11$$

Subtract 11 from both sides.

$$2x^{2}-12x-11=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -12 for b, and -11 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\left(-11\right)}}{2\times 2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 2\left(-11\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-12\right)±\sqrt{144-8\left(-11\right)}}{2\times 2}$$

Multiply -8 times -11.

$$x=\frac{-\left(-12\right)±\sqrt{144+88}}{2\times 2}$$

Add 144 to 88.

$$x=\frac{-\left(-12\right)±\sqrt{232}}{2\times 2}$$

Take the square root of 232.

$$x=\frac{-\left(-12\right)±2\sqrt{58}}{2\times 2}$$

The opposite of -12 is 12.

$$x=\frac{12±2\sqrt{58}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{12±2\sqrt{58}}{4}$$

Now solve the equation x=\frac{12±2\sqrt{58}}{4} when ± is plus. Add 12 to 2\sqrt{58}.

$$x=\frac{2\sqrt{58}+12}{4}$$

Divide 12+2\sqrt{58} by 4.

$$x=\frac{\sqrt{58}}{2}+3$$

Now solve the equation x=\frac{12±2\sqrt{58}}{4} when ± is minus. Subtract 2\sqrt{58} from 12.

$$x=\frac{12-2\sqrt{58}}{4}$$

Divide 12-2\sqrt{58} by 4.

$$x=-\frac{\sqrt{58}}{2}+3$$

The equation is now solved.

$$x=\frac{\sqrt{58}}{2}+3 x=-\frac{\sqrt{58}}{2}+3$$

Steps for Completing the Square

Combine -8x and -4x to get -12x.

$$2x^{2}-12x=11$$

Divide both sides by 2.

$$\frac{2x^{2}-12x}{2}=\frac{11}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{12}{2}\right)x=\frac{11}{2}$$

Divide -12 by 2.

$$x^{2}-6x=\frac{11}{2}$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=\frac{11}{2}+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=\frac{11}{2}+9$$

Add \frac{11}{2} to 9.

$$x^{2}-6x+9=\frac{29}{2}$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=\frac{29}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{\frac{29}{2}}$$

Simplify.

$$x-3=\frac{\sqrt{58}}{2} x-3=-\frac{\sqrt{58}}{2}$$

Add 3 to both sides of the equation.

$$x=\frac{\sqrt{58}}{2}+3 x=-\frac{\sqrt{58}}{2}+3$$