v^2-16v+52=0

Solution for v^2-16v+52=0 equation:

v^2-16v+52=0

a = 1; b = -16; c = +52;
Δ = b2-4ac
Δ = -162-4·1·52
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{3}}{2*1}=\frac{16-4\sqrt{3}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{3}}{2*1}=\frac{16+4\sqrt{3}}{2}
$