Solve 2x^2-4(x+2)=5x+2(2x-4)

$$2 x ^ { 2 } - 4 ( x + 2 ) = 5 x + 2 ( 2 x - 4 )$$

Solve for x

$$x = \frac{13}{2} = 6\frac{1}{2} = 6.5$$

$$x=0$$

Steps Using Factoring

Use the distributive property to multiply -4 by x+2.

$$2x^{2}-4x-8=5x+2\left(2x-4\right)$$

Use the distributive property to multiply 2 by 2x-4.

$$2x^{2}-4x-8=5x+4x-8$$

Combine 5x and 4x to get 9x.

$$2x^{2}-4x-8=9x-8$$

Subtract 9x from both sides.

$$2x^{2}-4x-8-9x=-8$$

Combine -4x and -9x to get -13x.

$$2x^{2}-13x-8=-8$$

Add 8 to both sides.

$$2x^{2}-13x-8+8=0$$

Add -8 and 8 to get 0.

$$2x^{2}-13x=0$$

Factor out x.

$$x\left(2x-13\right)=0$$

To find equation solutions, solve x=0 and 2x-13=0.

$$x=0 x=\frac{13}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -4 by x+2.

$$2x^{2}-4x-8=5x+2\left(2x-4\right)$$

Use the distributive property to multiply 2 by 2x-4.

$$2x^{2}-4x-8=5x+4x-8$$

Combine 5x and 4x to get 9x.

$$2x^{2}-4x-8=9x-8$$

Subtract 9x from both sides.

$$2x^{2}-4x-8-9x=-8$$

Combine -4x and -9x to get -13x.

$$2x^{2}-13x-8=-8$$

Add 8 to both sides.

$$2x^{2}-13x-8+8=0$$

Add -8 and 8 to get 0.

$$2x^{2}-13x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -13 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}}}{2\times 2}$$

Take the square root of \left(-13\right)^{2}.

$$x=\frac{-\left(-13\right)±13}{2\times 2}$$

The opposite of -13 is 13.

$$x=\frac{13±13}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{13±13}{4}$$

Now solve the equation x=\frac{13±13}{4} when ± is plus. Add 13 to 13.

$$x=\frac{26}{4}$$

Reduce the fraction \frac{26}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{13}{2}$$

Now solve the equation x=\frac{13±13}{4} when ± is minus. Subtract 13 from 13.

$$x=\frac{0}{4}$$

Divide 0 by 4.

$$x=0$$

The equation is now solved.

$$x=\frac{13}{2} x=0$$

Steps for Completing the Square

Use the distributive property to multiply -4 by x+2.

$$2x^{2}-4x-8=5x+2\left(2x-4\right)$$

Use the distributive property to multiply 2 by 2x-4.

$$2x^{2}-4x-8=5x+4x-8$$

Combine 5x and 4x to get 9x.

$$2x^{2}-4x-8=9x-8$$

Subtract 9x from both sides.

$$2x^{2}-4x-8-9x=-8$$

Combine -4x and -9x to get -13x.

$$2x^{2}-13x-8=-8$$

Add 8 to both sides.

$$2x^{2}-13x=-8+8$$

Add -8 and 8 to get 0.

$$2x^{2}-13x=0$$

Divide both sides by 2.

$$\frac{2x^{2}-13x}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{13}{2}x=\frac{0}{2}$$

Divide 0 by 2.

$$x^{2}-\frac{13}{2}x=0$$

Divide -\frac{13}{2}, the coefficient of the x term, by 2 to get -\frac{13}{4}. Then add the square of -\frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{2}x+\left(-\frac{13}{4}\right)^{2}=\left(-\frac{13}{4}\right)^{2}$$

Square -\frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{2}x+\frac{169}{16}=\frac{169}{16}$$

Factor x^{2}-\frac{13}{2}x+\frac{169}{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}.

$$\left(x-\frac{13}{4}\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{4}\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$x-\frac{13}{4}=\frac{13}{4} x-\frac{13}{4}=-\frac{13}{4}$$

Add \frac{13}{4} to both sides of the equation.

$$x=\frac{13}{2} x=0$$