Solve 3/4+y^2=1

$$\frac { 3 } { 4 } + y ^ { 2 } = 1$$

Solve for y

$$y=\frac{1}{2}=0.5$$

$$y=-\frac{1}{2}=-0.5$$

Steps Using Difference of Squares

Subtract 1 from both sides.

$$\frac{3}{4}+y^{2}-1=0$$

Subtract 1 from \frac{3}{4} to get -\frac{1}{4}.

$$-\frac{1}{4}+y^{2}=0$$

Multiply both sides by 4.

$$-1+4y^{2}=0$$

Consider -1+4y^{2}. Rewrite -1+4y^{2} as \left(2y\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2y-1\right)\left(2y+1\right)=0$$

To find equation solutions, solve 2y-1=0 and 2y+1=0.

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

Steps by Finding Square Root

Subtract \frac{3}{4} from both sides.

$$y^{2}=1-\frac{3}{4}$$

Subtract \frac{3}{4} from 1 to get \frac{1}{4}.

$$y^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Subtract 1 from both sides.

$$\frac{3}{4}+y^{2}-1=0$$

Subtract 1 from \frac{3}{4} to get -\frac{1}{4}.

$$-\frac{1}{4}+y^{2}=0$$

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.

$$y^{2}-\frac{1}{4}=0$$

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

$$y=\frac{0±\sqrt{0^{2}-4\left(-\frac{1}{4}\right)}}{2}$$

Square 0.

$$y=\frac{0±\sqrt{-4\left(-\frac{1}{4}\right)}}{2}$$

Multiply -4 times -\frac{1}{4}.

$$y=\frac{0±\sqrt{1}}{2}$$

Take the square root of 1.

$$y=\frac{0±1}{2}$$

Now solve the equation y=\frac{0±1}{2} when ± is plus. Divide 1 by 2.

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

Now solve the equation y=\frac{0±1}{2} when ± is minus. Divide -1 by 2.

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

The equation is now solved.

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