Solve 2v^2-15v+26=(v-4)^2

$$2 v ^ { 2 } - 15 v + 26 = ( v - 4 ) ^ { 2 }$$

Solve for v

$$v=2$$

$$v=5$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-4\right)^{2}.

$$2v^{2}-15v+26=v^{2}-8v+16$$

Subtract v^{2} from both sides.

$$2v^{2}-15v+26-v^{2}=-8v+16$$

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

$$v^{2}-15v+26=-8v+16$$

Add 8v to both sides.

$$v^{2}-15v+26+8v=16$$

Combine -15v and 8v to get -7v.

$$v^{2}-7v+26=16$$

Subtract 16 from both sides.

$$v^{2}-7v+26-16=0$$

Subtract 16 from 26 to get 10.

$$v^{2}-7v+10=0$$

To solve the equation, factor v^{2}-7v+10 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.

$$a+b=-7 ab=10$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -7.

$$a=-5 b=-2$$

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

$$\left(v-5\right)\left(v-2\right)$$

To find equation solutions, solve v-5=0 and v-2=0.

$$v=5 v=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-4\right)^{2}.

$$2v^{2}-15v+26=v^{2}-8v+16$$

Subtract v^{2} from both sides.

$$2v^{2}-15v+26-v^{2}=-8v+16$$

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

$$v^{2}-15v+26=-8v+16$$

Add 8v to both sides.

$$v^{2}-15v+26+8v=16$$

Combine -15v and 8v to get -7v.

$$v^{2}-7v+26=16$$

Subtract 16 from both sides.

$$v^{2}-7v+26-16=0$$

Subtract 16 from 26 to get 10.

$$v^{2}-7v+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as v^{2}+av+bv+10. To find a and b, set up a system to be solved.

$$a+b=-7 ab=1\times 10=10$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -7.

$$a=-5 b=-2$$

Rewrite v^{2}-7v+10 as \left(v^{2}-5v\right)+\left(-2v+10\right).

$$\left(v^{2}-5v\right)+\left(-2v+10\right)$$

Factor out v in the first and -2 in the second group.

$$v\left(v-5\right)-2\left(v-5\right)$$

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

$$\left(v-5\right)\left(v-2\right)$$

To find equation solutions, solve v-5=0 and v-2=0.

$$v=5 v=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-4\right)^{2}.

$$2v^{2}-15v+26=v^{2}-8v+16$$

Subtract v^{2} from both sides.

$$2v^{2}-15v+26-v^{2}=-8v+16$$

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

$$v^{2}-15v+26=-8v+16$$

Add 8v to both sides.

$$v^{2}-15v+26+8v=16$$

Combine -15v and 8v to get -7v.

$$v^{2}-7v+26=16$$

Subtract 16 from both sides.

$$v^{2}-7v+26-16=0$$

Subtract 16 from 26 to get 10.

$$v^{2}-7v+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$v=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 10}}{2}$$

Square -7.

$$v=\frac{-\left(-7\right)±\sqrt{49-4\times 10}}{2}$$

Multiply -4 times 10.

$$v=\frac{-\left(-7\right)±\sqrt{49-40}}{2}$$

Add 49 to -40.

$$v=\frac{-\left(-7\right)±\sqrt{9}}{2}$$

Take the square root of 9.

$$v=\frac{-\left(-7\right)±3}{2}$$

The opposite of -7 is 7.

$$v=\frac{7±3}{2}$$

Now solve the equation v=\frac{7±3}{2} when ± is plus. Add 7 to 3.

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

Divide 10 by 2.

$$v=5$$

Now solve the equation v=\frac{7±3}{2} when ± is minus. Subtract 3 from 7.

$$v=\frac{4}{2}$$

Divide 4 by 2.

$$v=2$$

The equation is now solved.

$$v=5 v=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(v-4\right)^{2}.

$$2v^{2}-15v+26=v^{2}-8v+16$$

Subtract v^{2} from both sides.

$$2v^{2}-15v+26-v^{2}=-8v+16$$

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

$$v^{2}-15v+26=-8v+16$$

Add 8v to both sides.

$$v^{2}-15v+26+8v=16$$

Combine -15v and 8v to get -7v.

$$v^{2}-7v+26=16$$

Subtract 26 from both sides.

$$v^{2}-7v=16-26$$

Subtract 26 from 16 to get -10.

$$v^{2}-7v=-10$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$v^{2}-7v+\left(-\frac{7}{2}\right)^{2}=-10+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$v^{2}-7v+\frac{49}{4}=-10+\frac{49}{4}$$

Add -10 to \frac{49}{4}.

$$v^{2}-7v+\frac{49}{4}=\frac{9}{4}$$

Factor v^{2}-7v+\frac{49}{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}.

$$\left(v-\frac{7}{2}\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(v-\frac{7}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

$$v-\frac{7}{2}=\frac{3}{2} v-\frac{7}{2}=-\frac{3}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$v=5 v=2$$