Solve y^2=16-12

$$y ^ { 2 } = 16 - 12$$

Solve for y

$$y=2$$

$$y=-2$$

Steps Using Difference of Squares

Subtract 12 from 16 to get 4.

$$y^{2}=4$$

Subtract 4 from both sides.

$$y^{2}-4=0$$

Consider y^{2}-4. Rewrite y^{2}-4 as y^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(y-2\right)\left(y+2\right)=0$$

To find equation solutions, solve y-2=0 and y+2=0.

$$y=2 y=-2$$

Steps by Finding Square Root

Subtract 12 from 16 to get 4.

$$y^{2}=4$$

Take the square root of both sides of the equation.

$$y=2 y=-2$$

Steps Using the Quadratic Formula

Subtract 12 from 16 to get 4.

$$y^{2}=4$$

Subtract 4 from both sides.

$$y^{2}-4=0$$

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

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

Square 0.

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

Multiply -4 times -4.

$$y=\frac{0±\sqrt{16}}{2}$$

Take the square root of 16.

$$y=\frac{0±4}{2}$$

Now solve the equation y=\frac{0±4}{2} when ± is plus. Divide 4 by 2.

$$y=2$$

Now solve the equation y=\frac{0±4}{2} when ± is minus. Divide -4 by 2.

$$y=-2$$

The equation is now solved.

$$y=2 y=-2$$