Solve 2v^2-10v+44=(v-7)^2

$$2 v ^ { 2 } - 10 v + 44 = ( v - 7 ) ^ { 2 }$$

Solve for v

$$v=-5$$

$$v=1$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-7\right)^{2}.

$$2v^{2}-10v+44=v^{2}-14v+49$$

Subtract v^{2} from both sides.

$$2v^{2}-10v+44-v^{2}=-14v+49$$

Combine 2v^{2} and -v^{2} to get v^{2}.

$$v^{2}-10v+44=-14v+49$$

Add 14v to both sides.

$$v^{2}-10v+44+14v=49$$

Combine -10v and 14v to get 4v.

$$v^{2}+4v+44=49$$

Subtract 49 from both sides.

$$v^{2}+4v+44-49=0$$

Subtract 49 from 44 to get -5.

$$v^{2}+4v-5=0$$

To solve the equation, factor v^{2}+4v-5 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.

$$a+b=4 ab=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=5$$

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

$$\left(v-1\right)\left(v+5\right)$$

To find equation solutions, solve v-1=0 and v+5=0.

$$v=1 v=-5$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-7\right)^{2}.

$$2v^{2}-10v+44=v^{2}-14v+49$$

Subtract v^{2} from both sides.

$$2v^{2}-10v+44-v^{2}=-14v+49$$

Combine 2v^{2} and -v^{2} to get v^{2}.

$$v^{2}-10v+44=-14v+49$$

Add 14v to both sides.

$$v^{2}-10v+44+14v=49$$

Combine -10v and 14v to get 4v.

$$v^{2}+4v+44=49$$

Subtract 49 from both sides.

$$v^{2}+4v+44-49=0$$

Subtract 49 from 44 to get -5.

$$v^{2}+4v-5=0$$

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

$$a+b=4 ab=1\left(-5\right)=-5$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=5$$

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

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

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

$$v\left(v-1\right)+5\left(v-1\right)$$

Factor out common term v-1 by using distributive property.

$$\left(v-1\right)\left(v+5\right)$$

To find equation solutions, solve v-1=0 and v+5=0.

$$v=1 v=-5$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-7\right)^{2}.

$$2v^{2}-10v+44=v^{2}-14v+49$$

Subtract v^{2} from both sides.

$$2v^{2}-10v+44-v^{2}=-14v+49$$

Combine 2v^{2} and -v^{2} to get v^{2}.

$$v^{2}-10v+44=-14v+49$$

Add 14v to both sides.

$$v^{2}-10v+44+14v=49$$

Combine -10v and 14v to get 4v.

$$v^{2}+4v+44=49$$

Subtract 49 from both sides.

$$v^{2}+4v+44-49=0$$

Subtract 49 from 44 to get -5.

$$v^{2}+4v-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 4 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

$$v=\frac{-4±\sqrt{16-4\left(-5\right)}}{2}$$

Multiply -4 times -5.

$$v=\frac{-4±\sqrt{16+20}}{2}$$

Add 16 to 20.

$$v=\frac{-4±\sqrt{36}}{2}$$

Take the square root of 36.

$$v=\frac{-4±6}{2}$$

Now solve the equation v=\frac{-4±6}{2} when ± is plus. Add -4 to 6.

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

Divide 2 by 2.

$$v=1$$

Now solve the equation v=\frac{-4±6}{2} when ± is minus. Subtract 6 from -4.

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

Divide -10 by 2.

$$v=-5$$

The equation is now solved.

$$v=1 v=-5$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-7\right)^{2}.

$$2v^{2}-10v+44=v^{2}-14v+49$$

Subtract v^{2} from both sides.

$$2v^{2}-10v+44-v^{2}=-14v+49$$

Combine 2v^{2} and -v^{2} to get v^{2}.

$$v^{2}-10v+44=-14v+49$$

Add 14v to both sides.

$$v^{2}-10v+44+14v=49$$

Combine -10v and 14v to get 4v.

$$v^{2}+4v+44=49$$

Subtract 44 from both sides.

$$v^{2}+4v=49-44$$

Subtract 44 from 49 to get 5.

$$v^{2}+4v=5$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$v^{2}+4v+2^{2}=5+2^{2}$$

Square 2.

$$v^{2}+4v+4=5+4$$

Add 5 to 4.

$$v^{2}+4v+4=9$$

Factor v^{2}+4v+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}.

$$\left(v+2\right)^{2}=9$$

Take the square root of both sides of the equation.

$$\sqrt{\left(v+2\right)^{2}}=\sqrt{9}$$

Simplify.

$$v+2=3 v+2=-3$$

Subtract 2 from both sides of the equation.

$$v=1 v=-5$$