Solve (2x-1)^2=5(2x-1)

$$( 2 x - 1 ) ^ { 2 } = 5 ( 2 x - 1 )$$

Solve for x

$$x=3$$

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

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

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

Use the distributive property to multiply 5 by 2x-1.

$$4x^{2}-4x+1=10x-5$$

Subtract 10x from both sides.

$$4x^{2}-4x+1-10x=-5$$

Combine -4x and -10x to get -14x.

$$4x^{2}-14x+1=-5$$

Add 5 to both sides.

$$4x^{2}-14x+1+5=0$$

Add 1 and 5 to get 6.

$$4x^{2}-14x+6=0$$

Divide both sides by 2.

$$2x^{2}-7x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-7 ab=2\times 3=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-6 b=-1$$

Rewrite 2x^{2}-7x+3 as \left(2x^{2}-6x\right)+\left(-x+3\right).

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

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

$$2x\left(x-3\right)-\left(x-3\right)$$

Factor out common term x-3 by using distributive property.

$$\left(x-3\right)\left(2x-1\right)$$

To find equation solutions, solve x-3=0 and 2x-1=0.

$$x=3 x=\frac{1}{2}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

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

Use the distributive property to multiply 5 by 2x-1.

$$4x^{2}-4x+1=10x-5$$

Subtract 10x from both sides.

$$4x^{2}-4x+1-10x=-5$$

Combine -4x and -10x to get -14x.

$$4x^{2}-14x+1=-5$$

Add 5 to both sides.

$$4x^{2}-14x+1+5=0$$

Add 1 and 5 to get 6.

$$4x^{2}-14x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -14 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 4\times 6}}{2\times 4}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 4\times 6}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-14\right)±\sqrt{196-16\times 6}}{2\times 4}$$

Multiply -16 times 6.

$$x=\frac{-\left(-14\right)±\sqrt{196-96}}{2\times 4}$$

Add 196 to -96.

$$x=\frac{-\left(-14\right)±\sqrt{100}}{2\times 4}$$

Take the square root of 100.

$$x=\frac{-\left(-14\right)±10}{2\times 4}$$

The opposite of -14 is 14.

$$x=\frac{14±10}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{14±10}{8}$$

Now solve the equation x=\frac{14±10}{8} when ± is plus. Add 14 to 10.

$$x=\frac{24}{8}$$

Divide 24 by 8.

$$x=3$$

Now solve the equation x=\frac{14±10}{8} when ± is minus. Subtract 10 from 14.

$$x=\frac{4}{8}$$

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

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

The equation is now solved.

$$x=3 x=\frac{1}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-1\right)^{2}.

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

Use the distributive property to multiply 5 by 2x-1.

$$4x^{2}-4x+1=10x-5$$

Subtract 10x from both sides.

$$4x^{2}-4x+1-10x=-5$$

Combine -4x and -10x to get -14x.

$$4x^{2}-14x+1=-5$$

Subtract 1 from both sides.

$$4x^{2}-14x=-5-1$$

Subtract 1 from -5 to get -6.

$$4x^{2}-14x=-6$$

Divide both sides by 4.

$$\frac{4x^{2}-14x}{4}=-\frac{6}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{14}{4}\right)x=-\frac{6}{4}$$

Reduce the fraction \frac{-14}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{7}{2}x=-\frac{6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{7}{2}x=-\frac{3}{2}$$

Divide -\frac{7}{2}, the coefficient of the x term, by 2 to get -\frac{7}{4}. Then add the square of -\frac{7}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{2}x+\left(-\frac{7}{4}\right)^{2}=-\frac{3}{2}+\left(-\frac{7}{4}\right)^{2}$$

Square -\frac{7}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=-\frac{3}{2}+\frac{49}{16}$$

Add -\frac{3}{2} to \frac{49}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{2}x+\frac{49}{16}=\frac{25}{16}$$

Factor x^{2}-\frac{7}{2}x+\frac{49}{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}.

$$\left(x-\frac{7}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{7}{4}=\frac{5}{4} x-\frac{7}{4}=-\frac{5}{4}$$

Add \frac{7}{4} to both sides of the equation.

$$x=3 x=\frac{1}{2}$$