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

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

Solve for x

$$x=\sqrt{3}\approx 1.732050808$$

$$x=-\sqrt{3}\approx -1.732050808$$

Steps by Finding Square Root

Use the distributive property to multiply 3 by x-2.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 3x from both sides.

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

Combine 3x and -3x to get 0.

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

Divide both sides by -2.

$$x^{2}=\frac{-6}{-2}$$

Divide -6 by -2 to get 3.

$$x^{2}=3$$

Take the square root of both sides of the equation.

$$x=\sqrt{3} x=-\sqrt{3}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3 by x-2.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 3x from both sides.

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

Combine 3x and -3x to get 0.

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

Add 6 to both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 0 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-2\right)\times 6}}{2\left(-2\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-2\right)\times 6}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{0±\sqrt{8\times 6}}{2\left(-2\right)}$$

Multiply 8 times 6.

$$x=\frac{0±\sqrt{48}}{2\left(-2\right)}$$

Take the square root of 48.

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

Multiply 2 times -2.

$$x=\frac{0±4\sqrt{3}}{-4}$$

Now solve the equation x=\frac{0±4\sqrt{3}}{-4} when ± is plus.

$$x=-\sqrt{3}$$

Now solve the equation x=\frac{0±4\sqrt{3}}{-4} when ± is minus.

$$x=\sqrt{3}$$

The equation is now solved.

$$x=-\sqrt{3} x=\sqrt{3}$$