Solve left(25x+5right)^2=100

$${ \left(25x+5 \right) }^{ 2 } = 100$$

Solve for x

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

$$x=-\frac{3}{5}=-0.6$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(25x+5\right)^{2}.

$$625x^{2}+250x+25=100$$

Subtract 100 from both sides.

$$625x^{2}+250x+25-100=0$$

Subtract 100 from 25 to get -75.

$$625x^{2}+250x-75=0$$

Divide both sides by 25.

$$25x^{2}+10x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 25x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=10 ab=25\left(-3\right)=-75$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -75.

$$-1,75 -3,25 -5,15$$

Calculate the sum for each pair.

$$-1+75=74 -3+25=22 -5+15=10$$

The solution is the pair that gives sum 10.

$$a=-5 b=15$$

Rewrite 25x^{2}+10x-3 as \left(25x^{2}-5x\right)+\left(15x-3\right).

$$\left(25x^{2}-5x\right)+\left(15x-3\right)$$

Factor out 5x in the first and 3 in the second group.

$$5x\left(5x-1\right)+3\left(5x-1\right)$$

Factor out common term 5x-1 by using distributive property.

$$\left(5x-1\right)\left(5x+3\right)$$

To find equation solutions, solve 5x-1=0 and 5x+3=0.

$$x=\frac{1}{5} x=-\frac{3}{5}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(25x+5\right)^{2}.

$$625x^{2}+250x+25=100$$

Subtract 100 from both sides.

$$625x^{2}+250x+25-100=0$$

Subtract 100 from 25 to get -75.

$$625x^{2}+250x-75=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 625 for a, 250 for b, and -75 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-250±\sqrt{250^{2}-4\times 625\left(-75\right)}}{2\times 625}$$

Square 250.

$$x=\frac{-250±\sqrt{62500-4\times 625\left(-75\right)}}{2\times 625}$$

Multiply -4 times 625.

$$x=\frac{-250±\sqrt{62500-2500\left(-75\right)}}{2\times 625}$$

Multiply -2500 times -75.

$$x=\frac{-250±\sqrt{62500+187500}}{2\times 625}$$

Add 62500 to 187500.

$$x=\frac{-250±\sqrt{250000}}{2\times 625}$$

Take the square root of 250000.

$$x=\frac{-250±500}{2\times 625}$$

Multiply 2 times 625.

$$x=\frac{-250±500}{1250}$$

Now solve the equation x=\frac{-250±500}{1250} when ± is plus. Add -250 to 500.

$$x=\frac{250}{1250}$$

Reduce the fraction \frac{250}{1250} to lowest terms by extracting and canceling out 250.

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

Now solve the equation x=\frac{-250±500}{1250} when ± is minus. Subtract 500 from -250.

$$x=-\frac{750}{1250}$$

Reduce the fraction \frac{-750}{1250} to lowest terms by extracting and canceling out 250.

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

The equation is now solved.

$$x=\frac{1}{5} x=-\frac{3}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(25x+5\right)^{2}.

$$625x^{2}+250x+25=100$$

Subtract 25 from both sides.

$$625x^{2}+250x=100-25$$

Subtract 25 from 100 to get 75.

$$625x^{2}+250x=75$$

Divide both sides by 625.

$$\frac{625x^{2}+250x}{625}=\frac{75}{625}$$

Dividing by 625 undoes the multiplication by 625.

$$x^{2}+\frac{250}{625}x=\frac{75}{625}$$

Reduce the fraction \frac{250}{625} to lowest terms by extracting and canceling out 125.

$$x^{2}+\frac{2}{5}x=\frac{75}{625}$$

Reduce the fraction \frac{75}{625} to lowest terms by extracting and canceling out 25.

$$x^{2}+\frac{2}{5}x=\frac{3}{25}$$

Divide \frac{2}{5}, the coefficient of the x term, by 2 to get \frac{1}{5}. Then add the square of \frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{5}x+\left(\frac{1}{5}\right)^{2}=\frac{3}{25}+\left(\frac{1}{5}\right)^{2}$$

Square \frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{3+1}{25}$$

Add \frac{3}{25} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{5}x+\frac{1}{25}=\frac{4}{25}$$

Factor x^{2}+\frac{2}{5}x+\frac{1}{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}.

$$\left(x+\frac{1}{5}\right)^{2}=\frac{4}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Simplify.

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

Subtract \frac{1}{5} from both sides of the equation.

$$x=\frac{1}{5} x=-\frac{3}{5}$$