Solve v^2-35=2v

$$v ^ { 2 } - 35 = 2 v$$

Solve for v

$$v=-5$$

$$v=7$$

Steps Using Factoring

Subtract 2v from both sides.

$$v^{2}-35-2v=0$$

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

$$v^{2}-2v-35=0$$

To solve the equation, factor v^{2}-2v-35 using formula v^{2}+\left(a+b\right)v+ab=\left(v+a\right)\left(v+b\right). To find a and b, set up a system to be solved.

$$a+b=-2 ab=-35$$

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 -35.

$$1,-35 5,-7$$

Calculate the sum for each pair.

$$1-35=-34 5-7=-2$$

The solution is the pair that gives sum -2.

$$a=-7 b=5$$

Rewrite factored expression \left(v+a\right)\left(v+b\right) using the obtained values.

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

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

$$v=7 v=-5$$

Steps Using Factoring By Grouping

Subtract 2v from both sides.

$$v^{2}-35-2v=0$$

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

$$v^{2}-2v-35=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as v^{2}+av+bv-35. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-35\right)=-35$$

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 -35.

$$1,-35 5,-7$$

Calculate the sum for each pair.

$$1-35=-34 5-7=-2$$

The solution is the pair that gives sum -2.

$$a=-7 b=5$$

Rewrite v^{2}-2v-35 as \left(v^{2}-7v\right)+\left(5v-35\right).

$$\left(v^{2}-7v\right)+\left(5v-35\right)$$

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

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

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

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

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

$$v=7 v=-5$$

Steps Using the Quadratic Formula

Subtract 2v from both sides.

$$v^{2}-35-2v=0$$

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.

$$v^{2}-2v-35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$v=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-35\right)}}{2}$$

Square -2.

$$v=\frac{-\left(-2\right)±\sqrt{4-4\left(-35\right)}}{2}$$

Multiply -4 times -35.

$$v=\frac{-\left(-2\right)±\sqrt{4+140}}{2}$$

Add 4 to 140.

$$v=\frac{-\left(-2\right)±\sqrt{144}}{2}$$

Take the square root of 144.

$$v=\frac{-\left(-2\right)±12}{2}$$

The opposite of -2 is 2.

$$v=\frac{2±12}{2}$$

Now solve the equation v=\frac{2±12}{2} when ± is plus. Add 2 to 12.

$$v=\frac{14}{2}$$

Divide 14 by 2.

$$v=7$$

Now solve the equation v=\frac{2±12}{2} when ± is minus. Subtract 12 from 2.

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

Divide -10 by 2.

$$v=-5$$

The equation is now solved.

$$v=7 v=-5$$

Steps for Completing the Square

Subtract 2v from both sides.

$$v^{2}-35-2v=0$$

Add 35 to both sides. Anything plus zero gives itself.

$$v^{2}-2v=35$$

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.

$$v^{2}-2v+1=35+1$$

Add 35 to 1.

$$v^{2}-2v+1=36$$

Factor v^{2}-2v+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}.

$$\left(v-1\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(v-1\right)^{2}}=\sqrt{36}$$

Simplify.

$$v-1=6 v-1=-6$$

Add 1 to both sides of the equation.

$$v=7 v=-5$$