Solve (4x+7)^2=(x-2)^2

$$( 4 x + 7 ) ^ { 2 } = ( x - 2 ) ^ { 2 }$$

Solve for x

$$x=-3$$

$$x=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+7\right)^{2}.

$$16x^{2}+56x+49=\left(x-2\right)^{2}$$

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

$$16x^{2}+56x+49=x^{2}-4x+4$$

Subtract x^{2} from both sides.

$$16x^{2}+56x+49-x^{2}=-4x+4$$

Combine 16x^{2} and -x^{2} to get 15x^{2}.

$$15x^{2}+56x+49=-4x+4$$

Add 4x to both sides.

$$15x^{2}+56x+49+4x=4$$

Combine 56x and 4x to get 60x.

$$15x^{2}+60x+49=4$$

Subtract 4 from both sides.

$$15x^{2}+60x+49-4=0$$

Subtract 4 from 49 to get 45.

$$15x^{2}+60x+45=0$$

Divide both sides by 15.

$$x^{2}+4x+3=0$$

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

$$a+b=4 ab=1\times 3=3$$

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.

$$a=1 b=3$$

Rewrite x^{2}+4x+3 as \left(x^{2}+x\right)+\left(3x+3\right).

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

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

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

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

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

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

$$x=-1 x=-3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+7\right)^{2}.

$$16x^{2}+56x+49=\left(x-2\right)^{2}$$

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

$$16x^{2}+56x+49=x^{2}-4x+4$$

Subtract x^{2} from both sides.

$$16x^{2}+56x+49-x^{2}=-4x+4$$

Combine 16x^{2} and -x^{2} to get 15x^{2}.

$$15x^{2}+56x+49=-4x+4$$

Add 4x to both sides.

$$15x^{2}+56x+49+4x=4$$

Combine 56x and 4x to get 60x.

$$15x^{2}+60x+49=4$$

Subtract 4 from both sides.

$$15x^{2}+60x+49-4=0$$

Subtract 4 from 49 to get 45.

$$15x^{2}+60x+45=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 15 for a, 60 for b, and 45 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-60±\sqrt{60^{2}-4\times 15\times 45}}{2\times 15}$$

Square 60.

$$x=\frac{-60±\sqrt{3600-4\times 15\times 45}}{2\times 15}$$

Multiply -4 times 15.

$$x=\frac{-60±\sqrt{3600-60\times 45}}{2\times 15}$$

Multiply -60 times 45.

$$x=\frac{-60±\sqrt{3600-2700}}{2\times 15}$$

Add 3600 to -2700.

$$x=\frac{-60±\sqrt{900}}{2\times 15}$$

Take the square root of 900.

$$x=\frac{-60±30}{2\times 15}$$

Multiply 2 times 15.

$$x=\frac{-60±30}{30}$$

Now solve the equation x=\frac{-60±30}{30} when ± is plus. Add -60 to 30.

$$x=-\frac{30}{30}$$

Divide -30 by 30.

$$x=-1$$

Now solve the equation x=\frac{-60±30}{30} when ± is minus. Subtract 30 from -60.

$$x=-\frac{90}{30}$$

Divide -90 by 30.

$$x=-3$$

The equation is now solved.

$$x=-1 x=-3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+7\right)^{2}.

$$16x^{2}+56x+49=\left(x-2\right)^{2}$$

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

$$16x^{2}+56x+49=x^{2}-4x+4$$

Subtract x^{2} from both sides.

$$16x^{2}+56x+49-x^{2}=-4x+4$$

Combine 16x^{2} and -x^{2} to get 15x^{2}.

$$15x^{2}+56x+49=-4x+4$$

Add 4x to both sides.

$$15x^{2}+56x+49+4x=4$$

Combine 56x and 4x to get 60x.

$$15x^{2}+60x+49=4$$

Subtract 49 from both sides.

$$15x^{2}+60x=4-49$$

Subtract 49 from 4 to get -45.

$$15x^{2}+60x=-45$$

Divide both sides by 15.

$$\frac{15x^{2}+60x}{15}=-\frac{45}{15}$$

Dividing by 15 undoes the multiplication by 15.

$$x^{2}+\frac{60}{15}x=-\frac{45}{15}$$

Divide 60 by 15.

$$x^{2}+4x=-\frac{45}{15}$$

Divide -45 by 15.

$$x^{2}+4x=-3$$

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

$$x^{2}+4x+2^{2}=-3+2^{2}$$

Square 2.

$$x^{2}+4x+4=-3+4$$

Add -3 to 4.

$$x^{2}+4x+4=1$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+2=1 x+2=-1$$

Subtract 2 from both sides of the equation.

$$x=-1 x=-3$$