2v^2+8v=17

Solution for 2v^2+8v=17 equation:

2v^2+8v=17

We move all terms to the left:

2v^2+8v-(17)=0

a = 2; b = 8; c = -17;
Δ = b2-4ac
Δ = 82-4·2·(-17)
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-10\sqrt{2}}{2*2}=\frac{-8-10\sqrt{2}}{4}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+10\sqrt{2}}{2*2}=\frac{-8+10\sqrt{2}}{4}
$