-v^2+24v+16=-6v^2

Solution for -v^2+24v+16=-6v^2 equation:

-v^2+24v+16=-6v^2

We move all terms to the left:

-v^2+24v+16-(-6v^2)=0

We add all the numbers together, and all the variables

-1v^2-(-6v^2)+24v+16=0

We get rid of parentheses

-1v^2+6v^2+24v+16=0

We add all the numbers together, and all the variables

5v^2+24v+16=0

a = 5; b = 24; c = +16;
Δ = b2-4ac
Δ = 242-4·5·16
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-16}{2*5}=\frac{-40}{10}
=-4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+16}{2*5}=\frac{-8}{10}
=-4/5
$