v^2-20v+43=-3

Solution for v^2-20v+43=-3 equation:

v^2-20v+43=-3

We move all terms to the left:

v^2-20v+43-(-3)=0

We add all the numbers together, and all the variables

v^2-20v+46=0

a = 1; b = -20; c = +46;
Δ = b2-4ac
Δ = -202-4·1·46
Δ = 216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-6\sqrt{6}}{2*1}=\frac{20-6\sqrt{6}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+6\sqrt{6}}{2*1}=\frac{20+6\sqrt{6}}{2}
$