Solve y_0^2=3x_{0}-3quad(x_{0}+2)*(x_{0}-1)+3x_0^2-3

$$y _ { 0 } ^ { 2 } = 3 x _ { 0 } - 3 \quad ( x _ { 0 } + 2 ) \cdot ( x _ { 0 } - 1 ) + 3 x _ { 0 } ^ { 2 } - 3$$

Solve for y_0

$$y_{0}=\sqrt{3}$$

$$y_{0}=-\sqrt{3}$$

Steps by Finding Square Root

Multiply -1 and 3 to get -3.

$$y_{0}^{2}=3x_{0}-3\left(x_{0}+2\right)\left(x_{0}-1\right)+3x_{0}^{2}-3$$

Use the distributive property to multiply -3 by x_{0}+2.

$$y_{0}^{2}=3x_{0}+\left(-3x_{0}-6\right)\left(x_{0}-1\right)+3x_{0}^{2}-3$$

Use the distributive property to multiply -3x_{0}-6 by x_{0}-1 and combine like terms.

$$y_{0}^{2}=3x_{0}-3x_{0}^{2}-3x_{0}+6+3x_{0}^{2}-3$$

Combine 3x_{0} and -3x_{0} to get 0.

$$y_{0}^{2}=-3x_{0}^{2}+6+3x_{0}^{2}-3$$

Combine -3x_{0}^{2} and 3x_{0}^{2} to get 0.

$$y_{0}^{2}=6-3$$

Subtract 3 from 6 to get 3.

$$y_{0}^{2}=3$$

Take the square root of both sides of the equation.

$$y_{0}=\sqrt{3} y_{0}=-\sqrt{3}$$

Steps Using the Quadratic Formula

Subtract 3x_{0}-3\left(x_{0}+2\right)\left(x_{0}-1\right) from both sides.

$$y_{0}^{2}-\left(3x_{0}-3\left(x_{0}+2\right)\left(x_{0}-1\right)\right)=3x_{0}^{2}-3$$

Subtract 3x_{0}^{2} from both sides.

$$y_{0}^{2}-\left(3x_{0}-3\left(x_{0}+2\right)\left(x_{0}-1\right)\right)-3x_{0}^{2}=-3$$

Add 3 to both sides.

$$y_{0}^{2}-\left(3x_{0}-3\left(x_{0}+2\right)\left(x_{0}-1\right)\right)-3x_{0}^{2}+3=0$$

Use the distributive property to multiply -3 by x_{0}+2.

$$y_{0}^{2}-\left(3x_{0}+\left(-3x_{0}-6\right)\left(x_{0}-1\right)\right)-3x_{0}^{2}+3=0$$

Use the distributive property to multiply -3x_{0}-6 by x_{0}-1 and combine like terms.

$$y_{0}^{2}-\left(3x_{0}-3x_{0}^{2}-3x_{0}+6\right)-3x_{0}^{2}+3=0$$

Combine 3x_{0} and -3x_{0} to get 0.

$$y_{0}^{2}-\left(-3x_{0}^{2}+6\right)-3x_{0}^{2}+3=0$$

To find the opposite of -3x_{0}^{2}+6, find the opposite of each term.

$$y_{0}^{2}+3x_{0}^{2}-6-3x_{0}^{2}+3=0$$

Combine 3x_{0}^{2} and -3x_{0}^{2} to get 0.

$$y_{0}^{2}-6+3=0$$

Add -6 and 3 to get -3.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 0.

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

Multiply -4 times -3.

$$y_{0}=\frac{0±\sqrt{12}}{2}$$

Take the square root of 12.

$$y_{0}=\frac{0±2\sqrt{3}}{2}$$

Now solve the equation y_{0}=\frac{0±2\sqrt{3}}{2} when ± is plus.

$$y_{0}=\sqrt{3}$$

Now solve the equation y_{0}=\frac{0±2\sqrt{3}}{2} when ± is minus.

$$y_{0}=-\sqrt{3}$$

The equation is now solved.

$$y_{0}=\sqrt{3} y_{0}=-\sqrt{3}$$

Solve for x_0 (complex solution)

$$x_{0}\in \mathrm{C}$$

$$y_{0}=-\sqrt{3}\text{ or }y_{0}=\sqrt{3}$$

Solve for x_0

$$x_{0}\in \mathrm{R}$$

$$|y_{0}|=\sqrt{3}$$