Solve 12=(1-3x)(1-3x)+(1+3x)(1+3x)

$$12 = ( 1 - 3 x ) ( 1 - 3 x ) + ( 1 + 3 x ) ( 1 + 3 x )$$

Solve for x

$$x=\frac{\sqrt{5}}{3}\approx 0.745355992$$

$$x=-\frac{\sqrt{5}}{3}\approx -0.745355992$$

Steps by Finding Square Root

Multiply 1-3x and 1-3x to get \left(1-3x\right)^{2}.

$$12=\left(1-3x\right)^{2}+\left(1+3x\right)\left(1+3x\right)$$

Multiply 1+3x and 1+3x to get \left(1+3x\right)^{2}.

$$12=\left(1-3x\right)^{2}+\left(1+3x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-3x\right)^{2}.

$$12=1-6x+9x^{2}+\left(1+3x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+3x\right)^{2}.

$$12=1-6x+9x^{2}+1+6x+9x^{2}$$

Add 1 and 1 to get 2.

$$12=2-6x+9x^{2}+6x+9x^{2}$$

Combine -6x and 6x to get 0.

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

Combine 9x^{2} and 9x^{2} to get 18x^{2}.

$$12=2+18x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$2+18x^{2}=12$$

Subtract 2 from both sides.

$$18x^{2}=12-2$$

Subtract 2 from 12 to get 10.

$$18x^{2}=10$$

Divide both sides by 18.

$$x^{2}=\frac{10}{18}$$

Reduce the fraction \frac{10}{18} to lowest terms by extracting and canceling out 2.

$$x^{2}=\frac{5}{9}$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Multiply 1-3x and 1-3x to get \left(1-3x\right)^{2}.

$$12=\left(1-3x\right)^{2}+\left(1+3x\right)\left(1+3x\right)$$

Multiply 1+3x and 1+3x to get \left(1+3x\right)^{2}.

$$12=\left(1-3x\right)^{2}+\left(1+3x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-3x\right)^{2}.

$$12=1-6x+9x^{2}+\left(1+3x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+3x\right)^{2}.

$$12=1-6x+9x^{2}+1+6x+9x^{2}$$

Add 1 and 1 to get 2.

$$12=2-6x+9x^{2}+6x+9x^{2}$$

Combine -6x and 6x to get 0.

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

Combine 9x^{2} and 9x^{2} to get 18x^{2}.

$$12=2+18x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$2+18x^{2}=12$$

Subtract 12 from both sides.

$$2+18x^{2}-12=0$$

Subtract 12 from 2 to get -10.

$$-10+18x^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$18x^{2}-10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, 0 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 18\left(-10\right)}}{2\times 18}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 18\left(-10\right)}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{0±\sqrt{-72\left(-10\right)}}{2\times 18}$$

Multiply -72 times -10.

$$x=\frac{0±\sqrt{720}}{2\times 18}$$

Take the square root of 720.

$$x=\frac{0±12\sqrt{5}}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{0±12\sqrt{5}}{36}$$

Now solve the equation x=\frac{0±12\sqrt{5}}{36} when ± is plus.

$$x=\frac{\sqrt{5}}{3}$$

Now solve the equation x=\frac{0±12\sqrt{5}}{36} when ± is minus.

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

The equation is now solved.

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