2u^2+8u=0

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Solution for 2u^2+8u=0 equation: 



2u^2+8u=0

a = 2; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·2·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*2}=\frac{-16}{4}
=-4
$


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

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