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v^{2}=\left(\sqrt{12-v}\right)^{2}
Square both sides of the equation.
v^{2}=12-v
Calculate \sqrt{12-v} to the power of 2 and get 12-v.
v^{2}-12=-v
Subtract 12 from both sides.
v^{2}-12+v=0
Add v to both sides.
v^{2}+v-12=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=1 ab=-12
To solve the equation, factor v^{2}+v-12 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.
-1,12 -2,6 -3,4
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. List all such integer pairs that give product -12.
-1+12=11 -2+6=4 -3+4=1
Calculate the sum for each pair.
a=-3 b=4
The solution is the pair that gives sum 1.
\left(v-3\right)\left(v+4\right)
Rewrite factored expression \left(v+a\right)\left(v+b\right) using the obtained values.
v=3 v=-4
To find equation solutions, solve v-3=0 and v+4=0.
3=\sqrt{12-3}
Substitute 3 for v in the equation v=\sqrt{12-v}.
3=3
Simplify. The value v=3 satisfies the equation.
-4=\sqrt{12-\left(-4\right)}
Substitute -4 for v in the equation v=\sqrt{12-v}.
-4=4
Simplify. The value v=-4 does not satisfy the equation because the left and the right hand side have opposite signs.
v=3
Equation v=\sqrt{12-v} has a unique solution.