Solve (2x-3)^2+2x-(3-4x)=x^2-6x+9

$$( 2 x - 3 ) ^ { 2 } + 2 x - ( 3 - 4 x ) = x ^ { 2 } - 6 x + 9$$

Solve for x

$$x=-1$$

$$x=1$$

Steps Using Difference of Squares

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

$$4x^{2}-12x+9+2x-\left(3-4x\right)=x^{2}-6x+9$$

Combine -12x and 2x to get -10x.

$$4x^{2}-10x+9-\left(3-4x\right)=x^{2}-6x+9$$

To find the opposite of 3-4x, find the opposite of each term.

$$4x^{2}-10x+9-3+4x=x^{2}-6x+9$$

Subtract 3 from 9 to get 6.

$$4x^{2}-10x+6+4x=x^{2}-6x+9$$

Combine -10x and 4x to get -6x.

$$4x^{2}-6x+6=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$4x^{2}-6x+6-x^{2}=-6x+9$$

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

$$3x^{2}-6x+6=-6x+9$$

Add 6x to both sides.

$$3x^{2}-6x+6+6x=9$$

Combine -6x and 6x to get 0.

$$3x^{2}+6=9$$

Subtract 9 from both sides.

$$3x^{2}+6-9=0$$

Subtract 9 from 6 to get -3.

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

Divide both sides by 3.

$$x^{2}-1=0$$

Consider x^{2}-1. Rewrite x^{2}-1 as x^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-1\right)\left(x+1\right)=0$$

To find equation solutions, solve x-1=0 and x+1=0.

$$x=1 x=-1$$

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}.

$$4x^{2}-12x+9+2x-\left(3-4x\right)=x^{2}-6x+9$$

Combine -12x and 2x to get -10x.

$$4x^{2}-10x+9-\left(3-4x\right)=x^{2}-6x+9$$

To find the opposite of 3-4x, find the opposite of each term.

$$4x^{2}-10x+9-3+4x=x^{2}-6x+9$$

Subtract 3 from 9 to get 6.

$$4x^{2}-10x+6+4x=x^{2}-6x+9$$

Combine -10x and 4x to get -6x.

$$4x^{2}-6x+6=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$4x^{2}-6x+6-x^{2}=-6x+9$$

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

$$3x^{2}-6x+6=-6x+9$$

Add 6x to both sides.

$$3x^{2}-6x+6+6x=9$$

Combine -6x and 6x to get 0.

$$3x^{2}+6=9$$

Subtract 6 from both sides.

$$3x^{2}=9-6$$

Subtract 6 from 9 to get 3.

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

Divide both sides by 3.

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

Divide 3 by 3 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}.

$$4x^{2}-12x+9+2x-\left(3-4x\right)=x^{2}-6x+9$$

Combine -12x and 2x to get -10x.

$$4x^{2}-10x+9-\left(3-4x\right)=x^{2}-6x+9$$

To find the opposite of 3-4x, find the opposite of each term.

$$4x^{2}-10x+9-3+4x=x^{2}-6x+9$$

Subtract 3 from 9 to get 6.

$$4x^{2}-10x+6+4x=x^{2}-6x+9$$

Combine -10x and 4x to get -6x.

$$4x^{2}-6x+6=x^{2}-6x+9$$

Subtract x^{2} from both sides.

$$4x^{2}-6x+6-x^{2}=-6x+9$$

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

$$3x^{2}-6x+6=-6x+9$$

Add 6x to both sides.

$$3x^{2}-6x+6+6x=9$$

Combine -6x and 6x to get 0.

$$3x^{2}+6=9$$

Subtract 9 from both sides.

$$3x^{2}+6-9=0$$

Subtract 9 from 6 to get -3.

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

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

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

Square 0.

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

Multiply -4 times 3.

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

Multiply -12 times -3.

$$x=\frac{0±\sqrt{36}}{2\times 3}$$

Take the square root of 36.

$$x=\frac{0±6}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{0±6}{6}$$

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

$$x=1$$

Now solve the equation x=\frac{0±6}{6} when ± is minus. Divide -6 by 6.

$$x=-1$$

The equation is now solved.

$$x=1 x=-1$$