Solve 761.3=684(1+r)^2

$$761.3 = 684 ( 1 + r ) ^ { 2 }$$

Solve for r

$$r=\frac{\sqrt{1446470}}{1140}-1\approx 0.054993695$$

$$r=-\frac{\sqrt{1446470}}{1140}-1\approx -2.054993695$$

Steps Using the Quadratic Formula

Divide both sides by 684.

$$\frac{761.3}{684}=\left(1+r\right)^{2}$$

Expand \frac{761.3}{684} by multiplying both numerator and the denominator by 10.

$$\frac{7613}{6840}=\left(1+r\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+r\right)^{2}.

$$\frac{7613}{6840}=1+2r+r^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$1+2r+r^{2}=\frac{7613}{6840}$$

Subtract \frac{7613}{6840} from both sides.

$$1+2r+r^{2}-\frac{7613}{6840}=0$$

Subtract \frac{7613}{6840} from 1 to get -\frac{773}{6840}.

$$-\frac{773}{6840}+2r+r^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$r^{2}+2r-\frac{773}{6840}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 2 for b, and -\frac{773}{6840} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-2±\sqrt{2^{2}-4\left(-\frac{773}{6840}\right)}}{2}$$

Square 2.

$$r=\frac{-2±\sqrt{4-4\left(-\frac{773}{6840}\right)}}{2}$$

Multiply -4 times -\frac{773}{6840}.

$$r=\frac{-2±\sqrt{4+\frac{773}{1710}}}{2}$$

Add 4 to \frac{773}{1710}.

$$r=\frac{-2±\sqrt{\frac{7613}{1710}}}{2}$$

Take the square root of \frac{7613}{1710}.

$$r=\frac{-2±\frac{\sqrt{1446470}}{570}}{2}$$

Now solve the equation r=\frac{-2±\frac{\sqrt{1446470}}{570}}{2} when ± is plus. Add -2 to \frac{\sqrt{1446470}}{570}.

$$r=\frac{\frac{\sqrt{1446470}}{570}-2}{2}$$

Divide -2+\frac{\sqrt{1446470}}{570} by 2.

$$r=\frac{\sqrt{1446470}}{1140}-1$$

Now solve the equation r=\frac{-2±\frac{\sqrt{1446470}}{570}}{2} when ± is minus. Subtract \frac{\sqrt{1446470}}{570} from -2.

$$r=\frac{-\frac{\sqrt{1446470}}{570}-2}{2}$$

Divide -2-\frac{\sqrt{1446470}}{570} by 2.

$$r=-\frac{\sqrt{1446470}}{1140}-1$$

The equation is now solved.

$$r=\frac{\sqrt{1446470}}{1140}-1 r=-\frac{\sqrt{1446470}}{1140}-1$$

Steps for Completing the Square

Divide both sides by 684.

$$\frac{761.3}{684}=\left(1+r\right)^{2}$$

Expand \frac{761.3}{684} by multiplying both numerator and the denominator by 10.

$$\frac{7613}{6840}=\left(1+r\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+r\right)^{2}.

$$\frac{7613}{6840}=1+2r+r^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$1+2r+r^{2}=\frac{7613}{6840}$$

Subtract 1 from both sides.

$$2r+r^{2}=\frac{7613}{6840}-1$$

Subtract 1 from \frac{7613}{6840} to get \frac{773}{6840}.

$$2r+r^{2}=\frac{773}{6840}$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$r^{2}+2r=\frac{773}{6840}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}+2r+1^{2}=\frac{773}{6840}+1^{2}$$

Square 1.

$$r^{2}+2r+1=\frac{773}{6840}+1$$

Add \frac{773}{6840} to 1.

$$r^{2}+2r+1=\frac{7613}{6840}$$

Factor r^{2}+2r+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r+1\right)^{2}=\frac{7613}{6840}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r+1\right)^{2}}=\sqrt{\frac{7613}{6840}}$$

Simplify.

$$r+1=\frac{\sqrt{1446470}}{1140} r+1=-\frac{\sqrt{1446470}}{1140}$$

Subtract 1 from both sides of the equation.

$$r=\frac{\sqrt{1446470}}{1140}-1 r=-\frac{\sqrt{1446470}}{1140}-1$$