4z^2+25z=0

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Solution for 4z^2+25z=0 equation: 



4z^2+25z=0

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

$\sqrt{\Delta}=\sqrt{625}=25$


$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*4}=\frac{-50}{8}
=-6+1/4
$


$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*4}=\frac{0}{8}
=0
$

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