2x^2+80x=0

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Solution for 2x^2+80x=0 equation: 



2x^2+80x=0

a = 2; b = 80; c = 0;
Δ = b2-4ac
Δ = 802-4·2·0
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$


$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(80)-80}{2*2}=\frac{-160}{4}
=-40
$


$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(80)+80}{2*2}=\frac{0}{4}
=0
$

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