Solve (2x+1)^2=3(x+3)^2

$$( 2 x + 1 ) ^ { 2 } = 3 ( x + 3 ) ^ { 2 }$$

Solve for x

$$x=5\sqrt{3}+7\approx 15.660254038$$

$$x=7-5\sqrt{3}\approx -1.660254038$$

Steps Using the Quadratic Formula

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

$$4x^{2}+4x+1=3\left(x+3\right)^{2}$$

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

$$4x^{2}+4x+1=3\left(x^{2}+6x+9\right)$$

Use the distributive property to multiply 3 by x^{2}+6x+9.

$$4x^{2}+4x+1=3x^{2}+18x+27$$

Subtract 3x^{2} from both sides.

$$4x^{2}+4x+1-3x^{2}=18x+27$$

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

$$x^{2}+4x+1=18x+27$$

Subtract 18x from both sides.

$$x^{2}+4x+1-18x=27$$

Combine 4x and -18x to get -14x.

$$x^{2}-14x+1=27$$

Subtract 27 from both sides.

$$x^{2}-14x+1-27=0$$

Subtract 27 from 1 to get -26.

$$x^{2}-14x-26=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -14 for b, and -26 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\left(-26\right)}}{2}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\left(-26\right)}}{2}$$

Multiply -4 times -26.

$$x=\frac{-\left(-14\right)±\sqrt{196+104}}{2}$$

Add 196 to 104.

$$x=\frac{-\left(-14\right)±\sqrt{300}}{2}$$

Take the square root of 300.

$$x=\frac{-\left(-14\right)±10\sqrt{3}}{2}$$

The opposite of -14 is 14.

$$x=\frac{14±10\sqrt{3}}{2}$$

Now solve the equation x=\frac{14±10\sqrt{3}}{2} when ± is plus. Add 14 to 10\sqrt{3}.

$$x=\frac{10\sqrt{3}+14}{2}$$

Divide 14+10\sqrt{3} by 2.

$$x=5\sqrt{3}+7$$

Now solve the equation x=\frac{14±10\sqrt{3}}{2} when ± is minus. Subtract 10\sqrt{3} from 14.

$$x=\frac{14-10\sqrt{3}}{2}$$

Divide 14-10\sqrt{3} by 2.

$$x=7-5\sqrt{3}$$

The equation is now solved.

$$x=5\sqrt{3}+7 x=7-5\sqrt{3}$$

Steps for Completing the Square

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

$$4x^{2}+4x+1=3\left(x+3\right)^{2}$$

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

$$4x^{2}+4x+1=3\left(x^{2}+6x+9\right)$$

Use the distributive property to multiply 3 by x^{2}+6x+9.

$$4x^{2}+4x+1=3x^{2}+18x+27$$

Subtract 3x^{2} from both sides.

$$4x^{2}+4x+1-3x^{2}=18x+27$$

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

$$x^{2}+4x+1=18x+27$$

Subtract 18x from both sides.

$$x^{2}+4x+1-18x=27$$

Combine 4x and -18x to get -14x.

$$x^{2}-14x+1=27$$

Subtract 1 from both sides.

$$x^{2}-14x=27-1$$

Subtract 1 from 27 to get 26.

$$x^{2}-14x=26$$

Divide -14, the coefficient of the x term, by 2 to get -7. Then add the square of -7 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-14x+\left(-7\right)^{2}=26+\left(-7\right)^{2}$$

Square -7.

$$x^{2}-14x+49=26+49$$

Add 26 to 49.

$$x^{2}-14x+49=75$$

Factor x^{2}-14x+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}.

$$\left(x-7\right)^{2}=75$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-7\right)^{2}}=\sqrt{75}$$

Simplify.

$$x-7=5\sqrt{3} x-7=-5\sqrt{3}$$

Add 7 to both sides of the equation.

$$x=5\sqrt{3}+7 x=7-5\sqrt{3}$$