Solve 2(x^2-5x)=6x-8

$$2 ( x ^ { 2 } - 5 x ) = 6 x - 8$$

Solve for x

$$x=2\sqrt{3}+4\approx 7.464101615$$

$$x=4-2\sqrt{3}\approx 0.535898385$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by x^{2}-5x.

$$2x^{2}-10x=6x-8$$

Subtract 6x from both sides.

$$2x^{2}-10x-6x=-8$$

Combine -10x and -6x to get -16x.

$$2x^{2}-16x=-8$$

Add 8 to both sides.

$$2x^{2}-16x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -16 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 2\times 8}}{2\times 2}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 2\times 8}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-16\right)±\sqrt{256-8\times 8}}{2\times 2}$$

Multiply -8 times 8.

$$x=\frac{-\left(-16\right)±\sqrt{256-64}}{2\times 2}$$

Add 256 to -64.

$$x=\frac{-\left(-16\right)±\sqrt{192}}{2\times 2}$$

Take the square root of 192.

$$x=\frac{-\left(-16\right)±8\sqrt{3}}{2\times 2}$$

The opposite of -16 is 16.

$$x=\frac{16±8\sqrt{3}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{16±8\sqrt{3}}{4}$$

Now solve the equation x=\frac{16±8\sqrt{3}}{4} when ± is plus. Add 16 to 8\sqrt{3}.

$$x=\frac{8\sqrt{3}+16}{4}$$

Divide 16+8\sqrt{3} by 4.

$$x=2\sqrt{3}+4$$

Now solve the equation x=\frac{16±8\sqrt{3}}{4} when ± is minus. Subtract 8\sqrt{3} from 16.

$$x=\frac{16-8\sqrt{3}}{4}$$

Divide 16-8\sqrt{3} by 4.

$$x=4-2\sqrt{3}$$

The equation is now solved.

$$x=2\sqrt{3}+4 x=4-2\sqrt{3}$$

Steps for Completing the Square

Use the distributive property to multiply 2 by x^{2}-5x.

$$2x^{2}-10x=6x-8$$

Subtract 6x from both sides.

$$2x^{2}-10x-6x=-8$$

Combine -10x and -6x to get -16x.

$$2x^{2}-16x=-8$$

Divide both sides by 2.

$$\frac{2x^{2}-16x}{2}=-\frac{8}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{16}{2}\right)x=-\frac{8}{2}$$

Divide -16 by 2.

$$x^{2}-8x=-\frac{8}{2}$$

Divide -8 by 2.

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

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -4.

$$x^{2}-8x+16=-4+16$$

Add -4 to 16.

$$x^{2}-8x+16=12$$

Factor x^{2}-8x+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}.

$$\left(x-4\right)^{2}=12$$

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=2\sqrt{3} x-4=-2\sqrt{3}$$

Add 4 to both sides of the equation.

$$x=2\sqrt{3}+4 x=4-2\sqrt{3}$$