-v(v+4)+5=4v+1-5v

Solution for -v(v+4)+5=4v+1-5v equation:

-v(v+4)+5=4v+1-5v

We move all terms to the left:

-v(v+4)+5-(4v+1-5v)=0

We add all the numbers together, and all the variables

-v(v+4)-(-1v+1)+5=0

We multiply parentheses

-v^2-4v-(-1v+1)+5=0

We get rid of parentheses

-v^2-4v+1v-1+5=0

We add all the numbers together, and all the variables

-1v^2-3v+4=0

a = -1; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-1)·4
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*-1}=\frac{-2}{-2}
=1
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*-1}=\frac{8}{-2}
=-4
$