Solve 16+3y^2=57

$$16 + 3 y ^ { 2 } = 57$$

Solve for y

$$y = \frac{\sqrt{123}}{3} \approx 3.696845502$$

$$y = -\frac{\sqrt{123}}{3} \approx -3.696845502$$

Steps by Finding Square Root

Subtract 16 from both sides.

$$3y^{2}=57-16$$

Subtract 16 from 57 to get 41.

$$3y^{2}=41$$

Divide both sides by 3.

$$y^{2}=\frac{41}{3}$$

Take the square root of both sides of the equation.

$$y=\frac{\sqrt{123}}{3} y=-\frac{\sqrt{123}}{3}$$

Steps Using the Quadratic Formula

Subtract 57 from both sides.

$$16+3y^{2}-57=0$$

Subtract 57 from 16 to get -41.

$$-41+3y^{2}=0$$

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.

$$3y^{2}-41=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 0 for b, and -41 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\times 3\left(-41\right)}}{2\times 3}$$

Square 0.

$$y=\frac{0±\sqrt{-4\times 3\left(-41\right)}}{2\times 3}$$

Multiply -4 times 3.

$$y=\frac{0±\sqrt{-12\left(-41\right)}}{2\times 3}$$

Multiply -12 times -41.

$$y=\frac{0±\sqrt{492}}{2\times 3}$$

Take the square root of 492.

$$y=\frac{0±2\sqrt{123}}{2\times 3}$$

Multiply 2 times 3.

$$y=\frac{0±2\sqrt{123}}{6}$$

Now solve the equation y=\frac{0±2\sqrt{123}}{6} when ± is plus.

$$y=\frac{\sqrt{123}}{3}$$

Now solve the equation y=\frac{0±2\sqrt{123}}{6} when ± is minus.

$$y=-\frac{\sqrt{123}}{3}$$

The equation is now solved.

$$y=\frac{\sqrt{123}}{3} y=-\frac{\sqrt{123}}{3}$$