Solve (-5x+4)^2-4=0

$$( - 5 x + 4 ) ^ { 2 } - 4 = 0$$

Solve for x

$$x=\frac{2}{5}=0.4$$

$$x = \frac{6}{5} = 1\frac{1}{5} = 1.2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-5x+4\right)^{2}.

$$25x^{2}-40x+16-4=0$$

Subtract 4 from 16 to get 12.

$$25x^{2}-40x+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

$$a+b=-40 ab=25\times 12=300$$

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 300.

$$-1,-300 -2,-150 -3,-100 -4,-75 -5,-60 -6,-50 -10,-30 -12,-25 -15,-20$$

Calculate the sum for each pair.

$$-1-300=-301 -2-150=-152 -3-100=-103 -4-75=-79 -5-60=-65 -6-50=-56 -10-30=-40 -12-25=-37 -15-20=-35$$

The solution is the pair that gives sum -40.

$$a=-30 b=-10$$

Rewrite 25x^{2}-40x+12 as \left(25x^{2}-30x\right)+\left(-10x+12\right).

$$\left(25x^{2}-30x\right)+\left(-10x+12\right)$$

Factor out 5x in the first and -2 in the second group.

$$5x\left(5x-6\right)-2\left(5x-6\right)$$

Factor out common term 5x-6 by using distributive property.

$$\left(5x-6\right)\left(5x-2\right)$$

To find equation solutions, solve 5x-6=0 and 5x-2=0.

$$x=\frac{6}{5} x=\frac{2}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-5x+4\right)^{2}.

$$25x^{2}-40x+16-4=0$$

Subtract 4 from 16 to get 12.

$$25x^{2}-40x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -40 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 25\times 12}}{2\times 25}$$

Square -40.

$$x=\frac{-\left(-40\right)±\sqrt{1600-4\times 25\times 12}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-\left(-40\right)±\sqrt{1600-100\times 12}}{2\times 25}$$

Multiply -100 times 12.

$$x=\frac{-\left(-40\right)±\sqrt{1600-1200}}{2\times 25}$$

Add 1600 to -1200.

$$x=\frac{-\left(-40\right)±\sqrt{400}}{2\times 25}$$

Take the square root of 400.

$$x=\frac{-\left(-40\right)±20}{2\times 25}$$

The opposite of -40 is 40.

$$x=\frac{40±20}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{40±20}{50}$$

Now solve the equation x=\frac{40±20}{50} when ± is plus. Add 40 to 20.

$$x=\frac{60}{50}$$

Reduce the fraction \frac{60}{50} to lowest terms by extracting and canceling out 10.

$$x=\frac{6}{5}$$

Now solve the equation x=\frac{40±20}{50} when ± is minus. Subtract 20 from 40.

$$x=\frac{20}{50}$$

Reduce the fraction \frac{20}{50} to lowest terms by extracting and canceling out 10.

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

The equation is now solved.

$$x=\frac{6}{5} x=\frac{2}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-5x+4\right)^{2}.

$$25x^{2}-40x+16-4=0$$

Subtract 4 from 16 to get 12.

$$25x^{2}-40x+12=0$$

Subtract 12 from both sides. Anything subtracted from zero gives its negation.

$$25x^{2}-40x=-12$$

Divide both sides by 25.

$$\frac{25x^{2}-40x}{25}=-\frac{12}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\left(-\frac{40}{25}\right)x=-\frac{12}{25}$$

Reduce the fraction \frac{-40}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{8}{5}x=-\frac{12}{25}$$

Divide -\frac{8}{5}, the coefficient of the x term, by 2 to get -\frac{4}{5}. Then add the square of -\frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{5}x+\left(-\frac{4}{5}\right)^{2}=-\frac{12}{25}+\left(-\frac{4}{5}\right)^{2}$$

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

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{-12+16}{25}$$

Add -\frac{12}{25} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{4}{25}$$

Factor x^{2}-\frac{8}{5}x+\frac{16}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{4}{5}=\frac{2}{5} x-\frac{4}{5}=-\frac{2}{5}$$

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

$$x=\frac{6}{5} x=\frac{2}{5}$$