Solve y^2-12-29=0

$$y ^ { 2 } - 12 - 29 = 0$$

Solve for y

$$y=\sqrt{41}\approx 6.403124237$$

$$y=-\sqrt{41}\approx -6.403124237$$

Steps by Finding Square Root

Subtract 29 from -12 to get -41.

$$y^{2}-41=0$$

Add 41 to both sides. Anything plus zero gives itself.

$$y^{2}=41$$

Take the square root of both sides of the equation.

$$y=\sqrt{41} y=-\sqrt{41}$$

Steps Using the Quadratic Formula

Subtract 29 from -12 to get -41.

$$y^{2}-41=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -41 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\left(-41\right)}}{2}$$

Square 0.

$$y=\frac{0±\sqrt{-4\left(-41\right)}}{2}$$

Multiply -4 times -41.

$$y=\frac{0±\sqrt{164}}{2}$$

Take the square root of 164.

$$y=\frac{0±2\sqrt{41}}{2}$$

Now solve the equation y=\frac{0±2\sqrt{41}}{2} when ± is plus.

$$y=\sqrt{41}$$

Now solve the equation y=\frac{0±2\sqrt{41}}{2} when ± is minus.

$$y=-\sqrt{41}$$

The equation is now solved.

$$y=\sqrt{41} y=-\sqrt{41}$$