Solve 0=8x^2-20x+8

$$0=8 { x }^{ 2 } -20x+8$$

Solve for x

$$x=\frac{1}{2}=0.5$$

$$x=2$$

Steps Using Factoring By Grouping

Swap sides so that all variable terms are on the left hand side.

$$8x^{2}-20x+8=0$$

Divide both sides by 4.

$$2x^{2}-5x+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-5 ab=2\times 2=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

Rewrite 2x^{2}-5x+2 as \left(2x^{2}-4x\right)+\left(-x+2\right).

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

Factor out 2x in the first and -1 in the second group.

$$2x\left(x-2\right)-\left(x-2\right)$$

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(2x-1\right)$$

To find equation solutions, solve x-2=0 and 2x-1=0.

$$x=2 x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$8x^{2}-20x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -20 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 8\times 8}}{2\times 8}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 8\times 8}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-20\right)±\sqrt{400-32\times 8}}{2\times 8}$$

Multiply -32 times 8.

$$x=\frac{-\left(-20\right)±\sqrt{400-256}}{2\times 8}$$

Add 400 to -256.

$$x=\frac{-\left(-20\right)±\sqrt{144}}{2\times 8}$$

Take the square root of 144.

$$x=\frac{-\left(-20\right)±12}{2\times 8}$$

The opposite of -20 is 20.

$$x=\frac{20±12}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{20±12}{16}$$

Now solve the equation x=\frac{20±12}{16} when ± is plus. Add 20 to 12.

$$x=\frac{32}{16}$$

Divide 32 by 16.

$$x=2$$

Now solve the equation x=\frac{20±12}{16} when ± is minus. Subtract 12 from 20.

$$x=\frac{8}{16}$$

Reduce the fraction \frac{8}{16} to lowest terms by extracting and canceling out 8.

$$x=\frac{1}{2}$$

The equation is now solved.

$$x=2 x=\frac{1}{2}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$8x^{2}-20x+8=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$8x^{2}-20x=-8$$

Divide both sides by 8.

$$\frac{8x^{2}-20x}{8}=-\frac{8}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{20}{8}\right)x=-\frac{8}{8}$$

Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{5}{2}x=-\frac{8}{8}$$

Divide -8 by 8.

$$x^{2}-\frac{5}{2}x=-1$$

Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=-1+\left(-\frac{5}{4}\right)^{2}$$

Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=-1+\frac{25}{16}$$

Add -1 to \frac{25}{16}.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{9}{16}$$

Factor x^{2}-\frac{5}{2}x+\frac{25}{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}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x-\frac{5}{4}=\frac{3}{4} x-\frac{5}{4}=-\frac{3}{4}$$

Add \frac{5}{4} to both sides of the equation.

$$x=2 x=\frac{1}{2}$$