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

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

Solve for x

$$x=\sqrt{3}\approx 1.732050808$$

$$x=-\sqrt{3}\approx -1.732050808$$

Steps by Finding Square Root

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

$$4x^{2}+12x+9=3\left(x+2\right)^{2}$$

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

$$4x^{2}+12x+9=3\left(x^{2}+4x+4\right)$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

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

Subtract 3x^{2} from both sides.

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

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

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

Subtract 12x from both sides.

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

Combine 12x and -12x to get 0.

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

Subtract 9 from both sides.

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

Subtract 9 from 12 to get 3.

$$x^{2}=3$$

Take the square root of both sides of the equation.

$$x=\sqrt{3} x=-\sqrt{3}$$

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}.

$$4x^{2}+12x+9=3\left(x+2\right)^{2}$$

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

$$4x^{2}+12x+9=3\left(x^{2}+4x+4\right)$$

Use the distributive property to multiply 3 by x^{2}+4x+4.

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

Subtract 3x^{2} from both sides.

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

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

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

Subtract 12x from both sides.

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

Combine 12x and -12x to get 0.

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

Subtract 12 from both sides.

$$x^{2}+9-12=0$$

Subtract 12 from 9 to get -3.

$$x^{2}-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-3\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-3\right)}}{2}$$

Multiply -4 times -3.

$$x=\frac{0±\sqrt{12}}{2}$$

Take the square root of 12.

$$x=\frac{0±2\sqrt{3}}{2}$$

Now solve the equation x=\frac{0±2\sqrt{3}}{2} when ± is plus.

$$x=\sqrt{3}$$

Now solve the equation x=\frac{0±2\sqrt{3}}{2} when ± is minus.

$$x=-\sqrt{3}$$

The equation is now solved.

$$x=\sqrt{3} x=-\sqrt{3}$$