v^2+10v-1140=0

Solution for v^2+10v-1140=0 equation:

v^2+10v-1140=0

a = 1; b = 10; c = -1140;
Δ = b2-4ac
Δ = 102-4·1·(-1140)
Δ = 4660
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:

$\sqrt{\Delta}=\sqrt{4660}=\sqrt{4*1165}=\sqrt{4}*\sqrt{1165}=2\sqrt{1165}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{1165}}{2*1}=\frac{-10-2\sqrt{1165}}{2}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{1165}}{2*1}=\frac{-10+2\sqrt{1165}}{2}
$