b2+25b=0

Solution for b2+25b=0 equation:

b2+25b=0

We add all the numbers together, and all the variables

b^2+25b=0

a = 1; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·1·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*1}=\frac{-50}{2}
=-25
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*1}=\frac{0}{2}
=0
$