Solve (2x+3)^2=(x-2)^2+5

$$( 2 x + 3 ) ^ { 2 } = ( x - 2 ) ^ { 2 } + 5$$

Solve for x

$$x = -\frac{16}{3} = -5\frac{1}{3} \approx -5.333333333$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$4x^{2}+12x+9=\left(x-2\right)^{2}+5$$

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

$$4x^{2}+12x+9=x^{2}-4x+4+5$$

Add 4 and 5 to get 9.

$$4x^{2}+12x+9=x^{2}-4x+9$$

Subtract x^{2} from both sides.

$$4x^{2}+12x+9-x^{2}=-4x+9$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+12x+9=-4x+9$$

Add 4x to both sides.

$$3x^{2}+12x+9+4x=9$$

Combine 12x and 4x to get 16x.

$$3x^{2}+16x+9=9$$

Subtract 9 from both sides.

$$3x^{2}+16x+9-9=0$$

Subtract 9 from 9 to get 0.

$$3x^{2}+16x=0$$

Factor out x.

$$x\left(3x+16\right)=0$$

To find equation solutions, solve x=0 and 3x+16=0.

$$x=0 x=-\frac{16}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$4x^{2}+12x+9=\left(x-2\right)^{2}+5$$

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

$$4x^{2}+12x+9=x^{2}-4x+4+5$$

Add 4 and 5 to get 9.

$$4x^{2}+12x+9=x^{2}-4x+9$$

Subtract x^{2} from both sides.

$$4x^{2}+12x+9-x^{2}=-4x+9$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+12x+9=-4x+9$$

Add 4x to both sides.

$$3x^{2}+12x+9+4x=9$$

Combine 12x and 4x to get 16x.

$$3x^{2}+16x+9=9$$

Subtract 9 from both sides.

$$3x^{2}+16x+9-9=0$$

Subtract 9 from 9 to get 0.

$$3x^{2}+16x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 16 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}}}{2\times 3}$$

Take the square root of 16^{2}.

$$x=\frac{-16±16}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-16±16}{6}$$

Now solve the equation x=\frac{-16±16}{6} when ± is plus. Add -16 to 16.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

Now solve the equation x=\frac{-16±16}{6} when ± is minus. Subtract 16 from -16.

$$x=-\frac{32}{6}$$

Reduce the fraction \frac{-32}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{16}{3}$$

The equation is now solved.

$$x=0 x=-\frac{16}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$4x^{2}+12x+9=\left(x-2\right)^{2}+5$$

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

$$4x^{2}+12x+9=x^{2}-4x+4+5$$

Add 4 and 5 to get 9.

$$4x^{2}+12x+9=x^{2}-4x+9$$

Subtract x^{2} from both sides.

$$4x^{2}+12x+9-x^{2}=-4x+9$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+12x+9=-4x+9$$

Add 4x to both sides.

$$3x^{2}+12x+9+4x=9$$

Combine 12x and 4x to get 16x.

$$3x^{2}+16x+9=9$$

Subtract 9 from both sides.

$$3x^{2}+16x=9-9$$

Subtract 9 from 9 to get 0.

$$3x^{2}+16x=0$$

Divide both sides by 3.

$$\frac{3x^{2}+16x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{16}{3}x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}+\frac{16}{3}x=0$$

Divide \frac{16}{3}, the coefficient of the x term, by 2 to get \frac{8}{3}. Then add the square of \frac{8}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{16}{3}x+\left(\frac{8}{3}\right)^{2}=\left(\frac{8}{3}\right)^{2}$$

Square \frac{8}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{16}{3}x+\frac{64}{9}=\frac{64}{9}$$

Factor x^{2}+\frac{16}{3}x+\frac{64}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{8}{3}\right)^{2}=\frac{64}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{8}{3}\right)^{2}}=\sqrt{\frac{64}{9}}$$

Simplify.

$$x+\frac{8}{3}=\frac{8}{3} x+\frac{8}{3}=-\frac{8}{3}$$

Subtract \frac{8}{3} from both sides of the equation.

$$x=0 x=-\frac{16}{3}$$