Solve 5v^2=31v+28

$$5 v ^ { 2 } = 31 v + 28$$

Solve for v

$$v=-\frac{4}{5}=-0.8$$

$$v=7$$

Steps Using Factoring By Grouping

Subtract 31v from both sides.

$$5v^{2}-31v=28$$

Subtract 28 from both sides.

$$5v^{2}-31v-28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5v^{2}+av+bv-28. To find a and b, set up a system to be solved.

$$a+b=-31 ab=5\left(-28\right)=-140$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -140.

$$1,-140 2,-70 4,-35 5,-28 7,-20 10,-14$$

Calculate the sum for each pair.

$$1-140=-139 2-70=-68 4-35=-31 5-28=-23 7-20=-13 10-14=-4$$

The solution is the pair that gives sum -31.

$$a=-35 b=4$$

Rewrite 5v^{2}-31v-28 as \left(5v^{2}-35v\right)+\left(4v-28\right).

$$\left(5v^{2}-35v\right)+\left(4v-28\right)$$

Factor out 5v in the first and 4 in the second group.

$$5v\left(v-7\right)+4\left(v-7\right)$$

Factor out common term v-7 by using distributive property.

$$\left(v-7\right)\left(5v+4\right)$$

To find equation solutions, solve v-7=0 and 5v+4=0.

$$v=7 v=-\frac{4}{5}$$

Steps Using the Quadratic Formula

Subtract 31v from both sides.

$$5v^{2}-31v=28$$

Subtract 28 from both sides.

$$5v^{2}-31v-28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -31 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$v=\frac{-\left(-31\right)±\sqrt{\left(-31\right)^{2}-4\times 5\left(-28\right)}}{2\times 5}$$

Square -31.

$$v=\frac{-\left(-31\right)±\sqrt{961-4\times 5\left(-28\right)}}{2\times 5}$$

Multiply -4 times 5.

$$v=\frac{-\left(-31\right)±\sqrt{961-20\left(-28\right)}}{2\times 5}$$

Multiply -20 times -28.

$$v=\frac{-\left(-31\right)±\sqrt{961+560}}{2\times 5}$$

Add 961 to 560.

$$v=\frac{-\left(-31\right)±\sqrt{1521}}{2\times 5}$$

Take the square root of 1521.

$$v=\frac{-\left(-31\right)±39}{2\times 5}$$

The opposite of -31 is 31.

$$v=\frac{31±39}{2\times 5}$$

Multiply 2 times 5.

$$v=\frac{31±39}{10}$$

Now solve the equation v=\frac{31±39}{10} when ± is plus. Add 31 to 39.

$$v=\frac{70}{10}$$

Divide 70 by 10.

$$v=7$$

Now solve the equation v=\frac{31±39}{10} when ± is minus. Subtract 39 from 31.

$$v=-\frac{8}{10}$$

Reduce the fraction \frac{-8}{10} to lowest terms by extracting and canceling out 2.

$$v=-\frac{4}{5}$$

The equation is now solved.

$$v=7 v=-\frac{4}{5}$$

Steps for Completing the Square

Subtract 31v from both sides.

$$5v^{2}-31v=28$$

Divide both sides by 5.

$$\frac{5v^{2}-31v}{5}=\frac{28}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$v^{2}-\frac{31}{5}v=\frac{28}{5}$$

Divide -\frac{31}{5}, the coefficient of the x term, by 2 to get -\frac{31}{10}. Then add the square of -\frac{31}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$v^{2}-\frac{31}{5}v+\left(-\frac{31}{10}\right)^{2}=\frac{28}{5}+\left(-\frac{31}{10}\right)^{2}$$

Square -\frac{31}{10} by squaring both the numerator and the denominator of the fraction.

$$v^{2}-\frac{31}{5}v+\frac{961}{100}=\frac{28}{5}+\frac{961}{100}$$

Add \frac{28}{5} to \frac{961}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$v^{2}-\frac{31}{5}v+\frac{961}{100}=\frac{1521}{100}$$

Factor v^{2}-\frac{31}{5}v+\frac{961}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(v-\frac{31}{10}\right)^{2}=\frac{1521}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(v-\frac{31}{10}\right)^{2}}=\sqrt{\frac{1521}{100}}$$

Simplify.

$$v-\frac{31}{10}=\frac{39}{10} v-\frac{31}{10}=-\frac{39}{10}$$

Add \frac{31}{10} to both sides of the equation.

$$v=7 v=-\frac{4}{5}$$