4v^2-22v+27=0

Solution for 4v^2-22v+27=0 equation:

4v^2-22v+27=0

a = 4; b = -22; c = +27;
Δ = b2-4ac
Δ = -222-4·4·27
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{13}}{2*4}=\frac{22-2\sqrt{13}}{8}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{13}}{2*4}=\frac{22+2\sqrt{13}}{8}
$