Solve left(5x+2right)^2-left(2x-1right)^2=0

$${ \left(5x+2 \right) }^{ 2 } - { \left(2x-1 \right) }^{ 2 } =0$$

Solve for x

$$x=-\frac{1}{7}\approx -0.142857143$$

$$x=-1$$

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}.

$$25x^{2}+20x+4-\left(2x-1\right)^{2}=0$$

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

$$25x^{2}+20x+4-\left(4x^{2}-4x+1\right)=0$$

To find the opposite of 4x^{2}-4x+1, find the opposite of each term.

$$25x^{2}+20x+4-4x^{2}+4x-1=0$$

Combine 25x^{2} and -4x^{2} to get 21x^{2}.

$$21x^{2}+20x+4+4x-1=0$$

Combine 20x and 4x to get 24x.

$$21x^{2}+24x+4-1=0$$

Subtract 1 from 4 to get 3.

$$21x^{2}+24x+3=0$$

Divide both sides by 3.

$$7x^{2}+8x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=8 ab=7\times 1=7$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=7$$

Rewrite 7x^{2}+8x+1 as \left(7x^{2}+x\right)+\left(7x+1\right).

$$\left(7x^{2}+x\right)+\left(7x+1\right)$$

Factor out x in 7x^{2}+x.

$$x\left(7x+1\right)+7x+1$$

Factor out common term 7x+1 by using distributive property.

$$\left(7x+1\right)\left(x+1\right)$$

To find equation solutions, solve 7x+1=0 and x+1=0.

$$x=-\frac{1}{7} x=-1$$

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}.

$$25x^{2}+20x+4-\left(2x-1\right)^{2}=0$$

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

$$25x^{2}+20x+4-\left(4x^{2}-4x+1\right)=0$$

To find the opposite of 4x^{2}-4x+1, find the opposite of each term.

$$25x^{2}+20x+4-4x^{2}+4x-1=0$$

Combine 25x^{2} and -4x^{2} to get 21x^{2}.

$$21x^{2}+20x+4+4x-1=0$$

Combine 20x and 4x to get 24x.

$$21x^{2}+24x+4-1=0$$

Subtract 1 from 4 to get 3.

$$21x^{2}+24x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 21 for a, 24 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}-4\times 21\times 3}}{2\times 21}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\times 21\times 3}}{2\times 21}$$

Multiply -4 times 21.

$$x=\frac{-24±\sqrt{576-84\times 3}}{2\times 21}$$

Multiply -84 times 3.

$$x=\frac{-24±\sqrt{576-252}}{2\times 21}$$

Add 576 to -252.

$$x=\frac{-24±\sqrt{324}}{2\times 21}$$

Take the square root of 324.

$$x=\frac{-24±18}{2\times 21}$$

Multiply 2 times 21.

$$x=\frac{-24±18}{42}$$

Now solve the equation x=\frac{-24±18}{42} when ± is plus. Add -24 to 18.

$$x=-\frac{6}{42}$$

Reduce the fraction \frac{-6}{42} to lowest terms by extracting and canceling out 6.

$$x=-\frac{1}{7}$$

Now solve the equation x=\frac{-24±18}{42} when ± is minus. Subtract 18 from -24.

$$x=-\frac{42}{42}$$

Divide -42 by 42.

$$x=-1$$

The equation is now solved.

$$x=-\frac{1}{7} x=-1$$

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}.

$$25x^{2}+20x+4-\left(2x-1\right)^{2}=0$$

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

$$25x^{2}+20x+4-\left(4x^{2}-4x+1\right)=0$$

To find the opposite of 4x^{2}-4x+1, find the opposite of each term.

$$25x^{2}+20x+4-4x^{2}+4x-1=0$$

Combine 25x^{2} and -4x^{2} to get 21x^{2}.

$$21x^{2}+20x+4+4x-1=0$$

Combine 20x and 4x to get 24x.

$$21x^{2}+24x+4-1=0$$

Subtract 1 from 4 to get 3.

$$21x^{2}+24x+3=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

$$21x^{2}+24x=-3$$

Divide both sides by 21.

$$\frac{21x^{2}+24x}{21}=-\frac{3}{21}$$

Dividing by 21 undoes the multiplication by 21.

$$x^{2}+\frac{24}{21}x=-\frac{3}{21}$$

Reduce the fraction \frac{24}{21} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{8}{7}x=-\frac{3}{21}$$

Reduce the fraction \frac{-3}{21} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{8}{7}x=-\frac{1}{7}$$

Divide \frac{8}{7}, the coefficient of the x term, by 2 to get \frac{4}{7}. Then add the square of \frac{4}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{7}x+\left(\frac{4}{7}\right)^{2}=-\frac{1}{7}+\left(\frac{4}{7}\right)^{2}$$

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

$$x^{2}+\frac{8}{7}x+\frac{16}{49}=-\frac{1}{7}+\frac{16}{49}$$

Add -\frac{1}{7} to \frac{16}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{7}x+\frac{16}{49}=\frac{9}{49}$$

Factor x^{2}+\frac{8}{7}x+\frac{16}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{4}{7}\right)^{2}=\frac{9}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{7}\right)^{2}}=\sqrt{\frac{9}{49}}$$

Simplify.

$$x+\frac{4}{7}=\frac{3}{7} x+\frac{4}{7}=-\frac{3}{7}$$

Subtract \frac{4}{7} from both sides of the equation.

$$x=-\frac{1}{7} x=-1$$