Solve 4y^2=9

$$4 y ^ { 2 } = 9$$

Solve for y

$$y = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$y = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps Using Difference of Squares

Divide both sides by 4.

$$y^{2}=\frac{9}{4}$$

Subtract \frac{9}{4} from both sides.

$$y^{2}-\frac{9}{4}=0$$

Multiply both sides by 4.

$$4y^{2}-9=0$$

Consider 4y^{2}-9. Rewrite 4y^{2}-9 as \left(2y\right)^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2y-3\right)\left(2y+3\right)=0$$

To find equation solutions, solve 2y-3=0 and 2y+3=0.

$$y=\frac{3}{2} y=-\frac{3}{2}$$

Steps by Finding Square Root

Divide both sides by 4.

$$y^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$y=\frac{3}{2} y=-\frac{3}{2}$$

Steps Using the Quadratic Formula

Divide both sides by 4.

$$y^{2}=\frac{9}{4}$$

Subtract \frac{9}{4} from both sides.

$$y^{2}-\frac{9}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{9}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{0±\sqrt{0^{2}-4\left(-\frac{9}{4}\right)}}{2}$$

Square 0.

$$y=\frac{0±\sqrt{-4\left(-\frac{9}{4}\right)}}{2}$$

Multiply -4 times -\frac{9}{4}.

$$y=\frac{0±\sqrt{9}}{2}$$

Take the square root of 9.

$$y=\frac{0±3}{2}$$

Now solve the equation y=\frac{0±3}{2} when ± is plus. Divide 3 by 2.

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

Now solve the equation y=\frac{0±3}{2} when ± is minus. Divide -3 by 2.

$$y=-\frac{3}{2}$$

The equation is now solved.

$$y=\frac{3}{2} y=-\frac{3}{2}$$