Solve r-3/5=8/r

$$\frac{ r-3 }{ 5 } = \frac{ 8 }{ r }$$

Solve for r

$$r=-5$$

$$r=8$$

Steps Using the Quadratic Formula

Variable r cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5r, the least common multiple of 5,r.

$$r\left(r-3\right)=5\times 8$$

Use the distributive property to multiply r by r-3.

$$r^{2}-3r=5\times 8$$

Multiply 5 and 8 to get 40.

$$r^{2}-3r=40$$

Subtract 40 from both sides.

$$r^{2}-3r-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-40\right)}}{2}$$

Square -3.

$$r=\frac{-\left(-3\right)±\sqrt{9-4\left(-40\right)}}{2}$$

Multiply -4 times -40.

$$r=\frac{-\left(-3\right)±\sqrt{9+160}}{2}$$

Add 9 to 160.

$$r=\frac{-\left(-3\right)±\sqrt{169}}{2}$$

Take the square root of 169.

$$r=\frac{-\left(-3\right)±13}{2}$$

The opposite of -3 is 3.

$$r=\frac{3±13}{2}$$

Now solve the equation r=\frac{3±13}{2} when ± is plus. Add 3 to 13.

$$r=\frac{16}{2}$$

Divide 16 by 2.

$$r=8$$

Now solve the equation r=\frac{3±13}{2} when ± is minus. Subtract 13 from 3.

$$r=-\frac{10}{2}$$

Divide -10 by 2.

$$r=-5$$

The equation is now solved.

$$r=8 r=-5$$

Steps for Completing the Square

Variable r cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5r, the least common multiple of 5,r.

$$r\left(r-3\right)=5\times 8$$

Use the distributive property to multiply r by r-3.

$$r^{2}-3r=5\times 8$$

Multiply 5 and 8 to get 40.

$$r^{2}-3r=40$$

Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$r^{2}-3r+\left(-\frac{3}{2}\right)^{2}=40+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$r^{2}-3r+\frac{9}{4}=40+\frac{9}{4}$$

Add 40 to \frac{9}{4}.

$$r^{2}-3r+\frac{9}{4}=\frac{169}{4}$$

Factor r^{2}-3r+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(r-\frac{3}{2}\right)^{2}=\frac{169}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(r-\frac{3}{2}\right)^{2}}=\sqrt{\frac{169}{4}}$$

Simplify.

$$r-\frac{3}{2}=\frac{13}{2} r-\frac{3}{2}=-\frac{13}{2}$$

Add \frac{3}{2} to both sides of the equation.

$$r=8 r=-5$$