Solve 2v^2-3v+27=(v+3)^2

$$2 v ^ { 2 } - 3 v + 27 = ( v + 3 ) ^ { 2 }$$

Solve for v

$$v=3$$

$$v=6$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(v+3\right)^{2}.

$$2v^{2}-3v+27=v^{2}+6v+9$$

Subtract v^{2} from both sides.

$$2v^{2}-3v+27-v^{2}=6v+9$$

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

$$v^{2}-3v+27=6v+9$$

Subtract 6v from both sides.

$$v^{2}-3v+27-6v=9$$

Combine -3v and -6v to get -9v.

$$v^{2}-9v+27=9$$

Subtract 9 from both sides.

$$v^{2}-9v+27-9=0$$

Subtract 9 from 27 to get 18.

$$v^{2}-9v+18=0$$

To solve the equation, factor v^{2}-9v+18 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.

$$a+b=-9 ab=18$$

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 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -9.

$$a=-6 b=-3$$

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

$$\left(v-6\right)\left(v-3\right)$$

To find equation solutions, solve v-6=0 and v-3=0.

$$v=6 v=3$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(v+3\right)^{2}.

$$2v^{2}-3v+27=v^{2}+6v+9$$

Subtract v^{2} from both sides.

$$2v^{2}-3v+27-v^{2}=6v+9$$

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

$$v^{2}-3v+27=6v+9$$

Subtract 6v from both sides.

$$v^{2}-3v+27-6v=9$$

Combine -3v and -6v to get -9v.

$$v^{2}-9v+27=9$$

Subtract 9 from both sides.

$$v^{2}-9v+27-9=0$$

Subtract 9 from 27 to get 18.

$$v^{2}-9v+18=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as v^{2}+av+bv+18. To find a and b, set up a system to be solved.

$$a+b=-9 ab=1\times 18=18$$

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 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -9.

$$a=-6 b=-3$$

Rewrite v^{2}-9v+18 as \left(v^{2}-6v\right)+\left(-3v+18\right).

$$\left(v^{2}-6v\right)+\left(-3v+18\right)$$

Factor out v in the first and -3 in the second group.

$$v\left(v-6\right)-3\left(v-6\right)$$

Factor out common term v-6 by using distributive property.

$$\left(v-6\right)\left(v-3\right)$$

To find equation solutions, solve v-6=0 and v-3=0.

$$v=6 v=3$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(v+3\right)^{2}.

$$2v^{2}-3v+27=v^{2}+6v+9$$

Subtract v^{2} from both sides.

$$2v^{2}-3v+27-v^{2}=6v+9$$

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

$$v^{2}-3v+27=6v+9$$

Subtract 6v from both sides.

$$v^{2}-3v+27-6v=9$$

Combine -3v and -6v to get -9v.

$$v^{2}-9v+27=9$$

Subtract 9 from both sides.

$$v^{2}-9v+27-9=0$$

Subtract 9 from 27 to get 18.

$$v^{2}-9v+18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -9 for b, and 18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$v=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 18}}{2}$$

Square -9.

$$v=\frac{-\left(-9\right)±\sqrt{81-4\times 18}}{2}$$

Multiply -4 times 18.

$$v=\frac{-\left(-9\right)±\sqrt{81-72}}{2}$$

Add 81 to -72.

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

Take the square root of 9.

$$v=\frac{-\left(-9\right)±3}{2}$$

The opposite of -9 is 9.

$$v=\frac{9±3}{2}$$

Now solve the equation v=\frac{9±3}{2} when ± is plus. Add 9 to 3.

$$v=\frac{12}{2}$$

Divide 12 by 2.

$$v=6$$

Now solve the equation v=\frac{9±3}{2} when ± is minus. Subtract 3 from 9.

$$v=\frac{6}{2}$$

Divide 6 by 2.

$$v=3$$

The equation is now solved.

$$v=6 v=3$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(v+3\right)^{2}.

$$2v^{2}-3v+27=v^{2}+6v+9$$

Subtract v^{2} from both sides.

$$2v^{2}-3v+27-v^{2}=6v+9$$

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

$$v^{2}-3v+27=6v+9$$

Subtract 6v from both sides.

$$v^{2}-3v+27-6v=9$$

Combine -3v and -6v to get -9v.

$$v^{2}-9v+27=9$$

Subtract 27 from both sides.

$$v^{2}-9v=9-27$$

Subtract 27 from 9 to get -18.

$$v^{2}-9v=-18$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$v^{2}-9v+\left(-\frac{9}{2}\right)^{2}=-18+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$v^{2}-9v+\frac{81}{4}=-18+\frac{81}{4}$$

Add -18 to \frac{81}{4}.

$$v^{2}-9v+\frac{81}{4}=\frac{9}{4}$$

Factor v^{2}-9v+\frac{81}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$v-\frac{9}{2}=\frac{3}{2} v-\frac{9}{2}=-\frac{3}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$v=6 v=3$$