v^2-13v+36=)

Solution for v^2-13v+36=) equation:

v^2-13v+36=)

We move all terms to the left:

v^2-13v+36-())=0

We add all the numbers together, and all the variables

v^2-13v=0

a = 1; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·1·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*1}=\frac{0}{2}
=0
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*1}=\frac{26}{2}
=13
$