-5v^2+11v-28=-7v^2-4-2v

Solution for -5v^2+11v-28=-7v^2-4-2v equation:

-5v^2+11v-28=-7v^2-4-2v

We move all terms to the left:

-5v^2+11v-28-(-7v^2-4-2v)=0

We get rid of parentheses

-5v^2+7v^2+2v+11v+4-28=0

We add all the numbers together, and all the variables

2v^2+13v-24=0

a = 2; b = 13; c = -24;
Δ = b2-4ac
Δ = 132-4·2·(-24)
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{-(13)-19}{2*2}=\frac{-32}{4}
=-8
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*2}=\frac{6}{4}
=1+1/2
$