4m^2+18m=0

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Solution for 4m^2+18m=0 equation: 



4m^2+18m=0

a = 4; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·4·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$


$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*4}=\frac{-36}{8}
=-4+1/2
$


$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*4}=\frac{0}{8}
=0
$

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