v^2+8v=19

Solution for v^2+8v=19 equation:

v^2+8v=19

We move all terms to the left:

v^2+8v-(19)=0

a = 1; b = 8; c = -19;
Δ = b2-4ac
Δ = 82-4·1·(-19)
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{35}}{2*1}=\frac{-8-2\sqrt{35}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{35}}{2*1}=\frac{-8+2\sqrt{35}}{2}
$