2v^2=-3v+5

Solution for 2v^2=-3v+5 equation:

2v^2=-3v+5

We move all terms to the left:

2v^2-(-3v+5)=0

We get rid of parentheses

2v^2+3v-5=0

a = 2; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·2·(-5)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*2}=\frac{-10}{4}
=-2+1/2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*2}=\frac{4}{4}
=1
$