2v^2-42=-5v

Solution for 2v^2-42=-5v equation:

2v^2-42=-5v

We move all terms to the left:

2v^2-42-(-5v)=0

We get rid of parentheses

2v^2+5v-42=0

a = 2; b = 5; c = -42;
Δ = b2-4ac
Δ = 52-4·2·(-42)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-19}{2*2}=\frac{-24}{4}
=-6
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+19}{2*2}=\frac{14}{4}
=3+1/2
$