Solve 54(1+x)(1+x)=1215

$$54(1+x)(1+x)=1215$$

Solve for x

$$x=\frac{3\sqrt{10}}{2}-1\approx 3.74341649$$

$$x=-\frac{3\sqrt{10}}{2}-1\approx -5.74341649$$

Steps Using the Quadratic Formula

Multiply 1+x and 1+x to get \left(1+x\right)^{2}.

$$54\left(1+x\right)^{2}=1215$$

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

$$54\left(1+2x+x^{2}\right)=1215$$

Use the distributive property to multiply 54 by 1+2x+x^{2}.

$$54+108x+54x^{2}=1215$$

Subtract 1215 from both sides.

$$54+108x+54x^{2}-1215=0$$

Subtract 1215 from 54 to get -1161.

$$-1161+108x+54x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$54x^{2}+108x-1161=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 54 for a, 108 for b, and -1161 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-108±\sqrt{108^{2}-4\times 54\left(-1161\right)}}{2\times 54}$$

Square 108.

$$x=\frac{-108±\sqrt{11664-4\times 54\left(-1161\right)}}{2\times 54}$$

Multiply -4 times 54.

$$x=\frac{-108±\sqrt{11664-216\left(-1161\right)}}{2\times 54}$$

Multiply -216 times -1161.

$$x=\frac{-108±\sqrt{11664+250776}}{2\times 54}$$

Add 11664 to 250776.

$$x=\frac{-108±\sqrt{262440}}{2\times 54}$$

Take the square root of 262440.

$$x=\frac{-108±162\sqrt{10}}{2\times 54}$$

Multiply 2 times 54.

$$x=\frac{-108±162\sqrt{10}}{108}$$

Now solve the equation x=\frac{-108±162\sqrt{10}}{108} when ± is plus. Add -108 to 162\sqrt{10}.

$$x=\frac{162\sqrt{10}-108}{108}$$

Divide -108+162\sqrt{10} by 108.

$$x=\frac{3\sqrt{10}}{2}-1$$

Now solve the equation x=\frac{-108±162\sqrt{10}}{108} when ± is minus. Subtract 162\sqrt{10} from -108.

$$x=\frac{-162\sqrt{10}-108}{108}$$

Divide -108-162\sqrt{10} by 108.

$$x=-\frac{3\sqrt{10}}{2}-1$$

The equation is now solved.

$$x=\frac{3\sqrt{10}}{2}-1 x=-\frac{3\sqrt{10}}{2}-1$$

Steps for Completing the Square

Multiply 1+x and 1+x to get \left(1+x\right)^{2}.

$$54\left(1+x\right)^{2}=1215$$

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

$$54\left(1+2x+x^{2}\right)=1215$$

Use the distributive property to multiply 54 by 1+2x+x^{2}.

$$54+108x+54x^{2}=1215$$

Subtract 54 from both sides.

$$108x+54x^{2}=1215-54$$

Subtract 54 from 1215 to get 1161.

$$108x+54x^{2}=1161$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$54x^{2}+108x=1161$$

Divide both sides by 54.

$$\frac{54x^{2}+108x}{54}=\frac{1161}{54}$$

Dividing by 54 undoes the multiplication by 54.

$$x^{2}+\frac{108}{54}x=\frac{1161}{54}$$

Divide 108 by 54.

$$x^{2}+2x=\frac{1161}{54}$$

Reduce the fraction \frac{1161}{54} to lowest terms by extracting and canceling out 27.

$$x^{2}+2x=\frac{43}{2}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=\frac{43}{2}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{43}{2}+1$$

Add \frac{43}{2} to 1.

$$x^{2}+2x+1=\frac{45}{2}$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=\frac{45}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{45}{2}}$$

Simplify.

$$x+1=\frac{3\sqrt{10}}{2} x+1=-\frac{3\sqrt{10}}{2}$$

Subtract 1 from both sides of the equation.

$$x=\frac{3\sqrt{10}}{2}-1 x=-\frac{3\sqrt{10}}{2}-1$$