Solve (5x-2)^2-8=7(5x-2)

$$( 5 x - 2 ) ^ { 2 } - 8 = 7 ( 5 x - 2 )$$

Solve for x

$$x=\frac{1}{5}=0.2$$

$$x=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-2\right)^{2}.

$$25x^{2}-20x+4-8=7\left(5x-2\right)$$

Subtract 8 from 4 to get -4.

$$25x^{2}-20x-4=7\left(5x-2\right)$$

Use the distributive property to multiply 7 by 5x-2.

$$25x^{2}-20x-4=35x-14$$

Subtract 35x from both sides.

$$25x^{2}-20x-4-35x=-14$$

Combine -20x and -35x to get -55x.

$$25x^{2}-55x-4=-14$$

Add 14 to both sides.

$$25x^{2}-55x-4+14=0$$

Add -4 and 14 to get 10.

$$25x^{2}-55x+10=0$$

Divide both sides by 5.

$$5x^{2}-11x+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-11 ab=5\times 2=10$$

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 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -11.

$$a=-10 b=-1$$

Rewrite 5x^{2}-11x+2 as \left(5x^{2}-10x\right)+\left(-x+2\right).

$$\left(5x^{2}-10x\right)+\left(-x+2\right)$$

Factor out 5x in the first and -1 in the second group.

$$5x\left(x-2\right)-\left(x-2\right)$$

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

$$\left(x-2\right)\left(5x-1\right)$$

To find equation solutions, solve x-2=0 and 5x-1=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-2\right)^{2}.

$$25x^{2}-20x+4-8=7\left(5x-2\right)$$

Subtract 8 from 4 to get -4.

$$25x^{2}-20x-4=7\left(5x-2\right)$$

Use the distributive property to multiply 7 by 5x-2.

$$25x^{2}-20x-4=35x-14$$

Subtract 35x from both sides.

$$25x^{2}-20x-4-35x=-14$$

Combine -20x and -35x to get -55x.

$$25x^{2}-55x-4=-14$$

Add 14 to both sides.

$$25x^{2}-55x-4+14=0$$

Add -4 and 14 to get 10.

$$25x^{2}-55x+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -55 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-55\right)±\sqrt{\left(-55\right)^{2}-4\times 25\times 10}}{2\times 25}$$

Square -55.

$$x=\frac{-\left(-55\right)±\sqrt{3025-4\times 25\times 10}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-\left(-55\right)±\sqrt{3025-100\times 10}}{2\times 25}$$

Multiply -100 times 10.

$$x=\frac{-\left(-55\right)±\sqrt{3025-1000}}{2\times 25}$$

Add 3025 to -1000.

$$x=\frac{-\left(-55\right)±\sqrt{2025}}{2\times 25}$$

Take the square root of 2025.

$$x=\frac{-\left(-55\right)±45}{2\times 25}$$

The opposite of -55 is 55.

$$x=\frac{55±45}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{55±45}{50}$$

Now solve the equation x=\frac{55±45}{50} when ± is plus. Add 55 to 45.

$$x=\frac{100}{50}$$

Divide 100 by 50.

$$x=2$$

Now solve the equation x=\frac{55±45}{50} when ± is minus. Subtract 45 from 55.

$$x=\frac{10}{50}$$

Reduce the fraction \frac{10}{50} to lowest terms by extracting and canceling out 10.

$$x=\frac{1}{5}$$

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-2\right)^{2}.

$$25x^{2}-20x+4-8=7\left(5x-2\right)$$

Subtract 8 from 4 to get -4.

$$25x^{2}-20x-4=7\left(5x-2\right)$$

Use the distributive property to multiply 7 by 5x-2.

$$25x^{2}-20x-4=35x-14$$

Subtract 35x from both sides.

$$25x^{2}-20x-4-35x=-14$$

Combine -20x and -35x to get -55x.

$$25x^{2}-55x-4=-14$$

Add 4 to both sides.

$$25x^{2}-55x=-14+4$$

Add -14 and 4 to get -10.

$$25x^{2}-55x=-10$$

Divide both sides by 25.

$$\frac{25x^{2}-55x}{25}=-\frac{10}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\left(-\frac{55}{25}\right)x=-\frac{10}{25}$$

Reduce the fraction \frac{-55}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{11}{5}x=-\frac{10}{25}$$

Reduce the fraction \frac{-10}{25} to lowest terms by extracting and canceling out 5.

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

Divide -\frac{11}{5}, the coefficient of the x term, by 2 to get -\frac{11}{10}. Then add the square of -\frac{11}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{5}x+\left(-\frac{11}{10}\right)^{2}=-\frac{2}{5}+\left(-\frac{11}{10}\right)^{2}$$

Square -\frac{11}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=-\frac{2}{5}+\frac{121}{100}$$

Add -\frac{2}{5} to \frac{121}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{5}x+\frac{121}{100}=\frac{81}{100}$$

Factor x^{2}-\frac{11}{5}x+\frac{121}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{10}\right)^{2}=\frac{81}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Simplify.

$$x-\frac{11}{10}=\frac{9}{10} x-\frac{11}{10}=-\frac{9}{10}$$

Add \frac{11}{10} to both sides of the equation.

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