Solve (4x+1)^2=7x^2+8x+5

$$( 4 x + 1 ) ^ { 2 } = 7 x ^ { 2 } + 8 x + 5$$

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using Difference of Squares

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1=7x^{2}+8x+5$$

Subtract 7x^{2} from both sides.

$$16x^{2}+8x+1-7x^{2}=8x+5$$

Combine 16x^{2} and -7x^{2} to get 9x^{2}.

$$9x^{2}+8x+1=8x+5$$

Subtract 8x from both sides.

$$9x^{2}+8x+1-8x=5$$

Combine 8x and -8x to get 0.

$$9x^{2}+1=5$$

Subtract 5 from both sides.

$$9x^{2}+1-5=0$$

Subtract 5 from 1 to get -4.

$$9x^{2}-4=0$$

Consider 9x^{2}-4. Rewrite 9x^{2}-4 as \left(3x\right)^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(3x-2\right)\left(3x+2\right)=0$$

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

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

Steps by Finding Square Root

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1=7x^{2}+8x+5$$

Subtract 7x^{2} from both sides.

$$16x^{2}+8x+1-7x^{2}=8x+5$$

Combine 16x^{2} and -7x^{2} to get 9x^{2}.

$$9x^{2}+8x+1=8x+5$$

Subtract 8x from both sides.

$$9x^{2}+8x+1-8x=5$$

Combine 8x and -8x to get 0.

$$9x^{2}+1=5$$

Subtract 1 from both sides.

$$9x^{2}=5-1$$

Subtract 1 from 5 to get 4.

$$9x^{2}=4$$

Divide both sides by 9.

$$x^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1=7x^{2}+8x+5$$

Subtract 7x^{2} from both sides.

$$16x^{2}+8x+1-7x^{2}=8x+5$$

Combine 16x^{2} and -7x^{2} to get 9x^{2}.

$$9x^{2}+8x+1=8x+5$$

Subtract 8x from both sides.

$$9x^{2}+8x+1-8x=5$$

Combine 8x and -8x to get 0.

$$9x^{2}+1=5$$

Subtract 5 from both sides.

$$9x^{2}+1-5=0$$

Subtract 5 from 1 to get -4.

$$9x^{2}-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 0 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 9\left(-4\right)}}{2\times 9}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 9\left(-4\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{0±\sqrt{-36\left(-4\right)}}{2\times 9}$$

Multiply -36 times -4.

$$x=\frac{0±\sqrt{144}}{2\times 9}$$

Take the square root of 144.

$$x=\frac{0±12}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{0±12}{18}$$

Now solve the equation x=\frac{0±12}{18} when ± is plus. Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{0±12}{18} when ± is minus. Reduce the fraction \frac{-12}{18} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

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