Solve 1-r^2=3/4

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

Solve for r

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

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

Steps by Finding Square Root

Subtract 1 from both sides.

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

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

$$-r^{2}=-\frac{1}{4}$$

Divide both sides by -1.

$$r^{2}=\frac{-\frac{1}{4}}{-1}$$

Express \frac{-\frac{1}{4}}{-1} as a single fraction.

$$r^{2}=\frac{-1}{4\left(-1\right)}$$

Cancel out -1 in both numerator and denominator.

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

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

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

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

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

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

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.

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

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}.

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

Square 0.

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

Multiply -4 times -1.

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

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

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

Take the square root of 1.

$$r=\frac{0±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$r=\frac{0±1}{-2}$$

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

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

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

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

The equation is now solved.

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