(v-25)(v+16)=0

Solution for (v-25)(v+16)=0 equation:

(v-25)(v+16)=0

We multiply parentheses ..

(+v^2+16v-25v-400)=0

We get rid of parentheses

v^2+16v-25v-400=0

We add all the numbers together, and all the variables

v^2-9v-400=0

a = 1; b = -9; c = -400;
Δ = b2-4ac
Δ = -92-4·1·(-400)
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-41}{2*1}=\frac{-32}{2}
=-16
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+41}{2*1}=\frac{50}{2}
=25
$