Solve 2x-(x+1)^2=x^2-9

$$2 x - ( x + 1 ) ^ { 2 } = x ^ { 2 } - 9$$

Solve for x

$$x=2$$

$$x=-2$$

Steps by Finding Square Root

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

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

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$2x-x^{2}-2x-1=x^{2}-9$$

Combine 2x and -2x to get 0.

$$-x^{2}-1=x^{2}-9$$

Subtract x^{2} from both sides.

$$-x^{2}-1-x^{2}=-9$$

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

$$-2x^{2}-1=-9$$

Add 1 to both sides.

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

Add -9 and 1 to get -8.

$$-2x^{2}=-8$$

Divide both sides by -2.

$$x^{2}=\frac{-8}{-2}$$

Divide -8 by -2 to get 4.

$$x^{2}=4$$

Take the square root of both sides of the equation.

$$x=2 x=-2$$

Steps Using the Quadratic Formula

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

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

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$2x-x^{2}-2x-1=x^{2}-9$$

Combine 2x and -2x to get 0.

$$-x^{2}-1=x^{2}-9$$

Subtract x^{2} from both sides.

$$-x^{2}-1-x^{2}=-9$$

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

$$-2x^{2}-1=-9$$

Add 9 to both sides.

$$-2x^{2}-1+9=0$$

Add -1 and 9 to get 8.

$$-2x^{2}+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 0 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-2\right)\times 8}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{0±\sqrt{8\times 8}}{2\left(-2\right)}$$

Multiply 8 times 8.

$$x=\frac{0±\sqrt{64}}{2\left(-2\right)}$$

Take the square root of 64.

$$x=\frac{0±8}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{0±8}{-4}$$

Now solve the equation x=\frac{0±8}{-4} when ± is plus. Divide 8 by -4.

$$x=-2$$

Now solve the equation x=\frac{0±8}{-4} when ± is minus. Divide -8 by -4.

$$x=2$$

The equation is now solved.

$$x=-2 x=2$$