Solve 2v^2-40=2v

$$2 v ^ { 2 } - 40 = 2 v$$

Solve for v

$$v=-4$$

$$v=5$$

Steps Using Factoring By Grouping

Subtract 2v from both sides.

$$2v^{2}-40-2v=0$$

Divide both sides by 2.

$$v^{2}-20-v=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$v^{2}-v-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as v^{2}+av+bv-20. To find a and b, set up a system to be solved.

$$a+b=-1 ab=1\left(-20\right)=-20$$

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 -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -1.

$$a=-5 b=4$$

Rewrite v^{2}-v-20 as \left(v^{2}-5v\right)+\left(4v-20\right).

$$\left(v^{2}-5v\right)+\left(4v-20\right)$$

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

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

Factor out common term v-5 by using distributive property.

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

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

$$v=5 v=-4$$

Steps Using the Quadratic Formula

Subtract 2v from both sides.

$$2v^{2}-40-2v=0$$

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.

$$2v^{2}-2v-40=0$$

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

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

Square -2.

$$v=\frac{-\left(-2\right)±\sqrt{4-4\times 2\left(-40\right)}}{2\times 2}$$

Multiply -4 times 2.

$$v=\frac{-\left(-2\right)±\sqrt{4-8\left(-40\right)}}{2\times 2}$$

Multiply -8 times -40.

$$v=\frac{-\left(-2\right)±\sqrt{4+320}}{2\times 2}$$

Add 4 to 320.

$$v=\frac{-\left(-2\right)±\sqrt{324}}{2\times 2}$$

Take the square root of 324.

$$v=\frac{-\left(-2\right)±18}{2\times 2}$$

The opposite of -2 is 2.

$$v=\frac{2±18}{2\times 2}$$

Multiply 2 times 2.

$$v=\frac{2±18}{4}$$

Now solve the equation v=\frac{2±18}{4} when ± is plus. Add 2 to 18.

$$v=\frac{20}{4}$$

Divide 20 by 4.

$$v=5$$

Now solve the equation v=\frac{2±18}{4} when ± is minus. Subtract 18 from 2.

$$v=-\frac{16}{4}$$

Divide -16 by 4.

$$v=-4$$

The equation is now solved.

$$v=5 v=-4$$

Steps for Completing the Square

Subtract 2v from both sides.

$$2v^{2}-40-2v=0$$

Add 40 to both sides. Anything plus zero gives itself.

$$2v^{2}-2v=40$$

Divide both sides by 2.

$$\frac{2v^{2}-2v}{2}=\frac{40}{2}$$

Dividing by 2 undoes the multiplication by 2.

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

Divide -2 by 2.

$$v^{2}-v=\frac{40}{2}$$

Divide 40 by 2.

$$v^{2}-v=20$$

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

$$v^{2}-v+\left(-\frac{1}{2}\right)^{2}=20+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$v^{2}-v+\frac{1}{4}=20+\frac{1}{4}$$

Add 20 to \frac{1}{4}.

$$v^{2}-v+\frac{1}{4}=\frac{81}{4}$$

Factor v^{2}-v+\frac{1}{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}.

$$\left(v-\frac{1}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(v-\frac{1}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$v-\frac{1}{2}=\frac{9}{2} v-\frac{1}{2}=-\frac{9}{2}$$

Add \frac{1}{2} to both sides of the equation.

$$v=5 v=-4$$