Solve 2(x-1)=-x^2+2x+3

$$2 ( x - 1 ) = - x ^ { 2 } + 2 x + 3$$

Solve for x

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

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

Steps by Finding Square Root

Use the distributive property to multiply 2 by x-1.

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

Add x^{2} to both sides.

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

Subtract 2x from both sides.

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

Combine 2x and -2x to get 0.

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

Add 2 to both sides.

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

Add 3 and 2 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 the distributive property to multiply 2 by x-1.

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

Add x^{2} to both sides.

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

Subtract 2x from both sides.

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

Combine 2x and -2x to get 0.

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

Subtract 3 from both sides.

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

Subtract 3 from -2 to get -5.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 0.

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

Multiply -4 times -5.

$$x=\frac{0±\sqrt{20}}{2}$$

Take the square root of 20.

$$x=\frac{0±2\sqrt{5}}{2}$$

Now solve the equation x=\frac{0±2\sqrt{5}}{2} when ± is plus.

$$x=\sqrt{5}$$

Now solve the equation x=\frac{0±2\sqrt{5}}{2} when ± is minus.

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

The equation is now solved.

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