-4b^2+20b=0

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Solution for -4b^2+20b=0 equation:



-4b^2+20b=0
a = -4; b = 20; c = 0;
Δ = b2-4ac
Δ = 202-4·(-4)·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-20}{2*-4}=\frac{-40}{-8} =+5 $
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+20}{2*-4}=\frac{0}{-8} =0 $

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