Solve (2x-1)^2-(x-2)^2=12

$$( 2 x - 1 ) ^ { 2 } - ( x - 2 ) ^ { 2 } = 12$$

Solve for x

$$x=\sqrt{5}\approx 2.236067977$$

$$x=-\sqrt{5}\approx -2.236067977$$

Steps by Finding Square Root

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

$$4x^{2}-4x+1-\left(x-2\right)^{2}=12$$

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

$$4x^{2}-4x+1-\left(x^{2}-4x+4\right)=12$$

To find the opposite of x^{2}-4x+4, find the opposite of each term.

$$4x^{2}-4x+1-x^{2}+4x-4=12$$

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

$$3x^{2}-4x+1+4x-4=12$$

Combine -4x and 4x to get 0.

$$3x^{2}+1-4=12$$

Subtract 4 from 1 to get -3.

$$3x^{2}-3=12$$

Add 3 to both sides.

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

Add 12 and 3 to get 15.

$$3x^{2}=15$$

Divide both sides by 3.

$$x^{2}=\frac{15}{3}$$

Divide 15 by 3 to get 5.

$$x^{2}=5$$

Take the square root of both sides of the equation.

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

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}.

$$4x^{2}-4x+1-\left(x-2\right)^{2}=12$$

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

$$4x^{2}-4x+1-\left(x^{2}-4x+4\right)=12$$

To find the opposite of x^{2}-4x+4, find the opposite of each term.

$$4x^{2}-4x+1-x^{2}+4x-4=12$$

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

$$3x^{2}-4x+1+4x-4=12$$

Combine -4x and 4x to get 0.

$$3x^{2}+1-4=12$$

Subtract 4 from 1 to get -3.

$$3x^{2}-3=12$$

Subtract 12 from both sides.

$$3x^{2}-3-12=0$$

Subtract 12 from -3 to get -15.

$$3x^{2}-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 3\left(-15\right)}}{2\times 3}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 3\left(-15\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{0±\sqrt{-12\left(-15\right)}}{2\times 3}$$

Multiply -12 times -15.

$$x=\frac{0±\sqrt{180}}{2\times 3}$$

Take the square root of 180.

$$x=\frac{0±6\sqrt{5}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±6\sqrt{5}}{6}$$

Now solve the equation x=\frac{0±6\sqrt{5}}{6} when ± is plus.

$$x=\sqrt{5}$$

Now solve the equation x=\frac{0±6\sqrt{5}}{6} when ± is minus.

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

The equation is now solved.

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