Solve left(2x-3right)^2=left(3x-2right)^2

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

Solve for x

$$x=1$$

$$x=-1$$

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}.

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

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

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

Subtract 9x^{2} from both sides.

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

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

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

Add 12x to both sides.

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

Combine -12x and 12x to get 0.

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

Subtract 9 from both sides.

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

Subtract 9 from 4 to get -5.

$$-5x^{2}=-5$$

Divide both sides by -5.

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

Divide -5 by -5 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

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}.

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

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

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

Subtract 9x^{2} from both sides.

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

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

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

Add 12x to both sides.

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

Combine -12x and 12x to get 0.

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

Subtract 4 from both sides.

$$-5x^{2}+9-4=0$$

Subtract 4 from 9 to get 5.

$$-5x^{2}+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 0 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-5\right)\times 5}}{2\left(-5\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-5\right)\times 5}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{0±\sqrt{20\times 5}}{2\left(-5\right)}$$

Multiply 20 times 5.

$$x=\frac{0±\sqrt{100}}{2\left(-5\right)}$$

Take the square root of 100.

$$x=\frac{0±10}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{0±10}{-10}$$

Now solve the equation x=\frac{0±10}{-10} when ± is plus. Divide 10 by -10.

$$x=-1$$

Now solve the equation x=\frac{0±10}{-10} when ± is minus. Divide -10 by -10.

$$x=1$$

The equation is now solved.

$$x=-1 x=1$$