5t2+40t=0

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Solution for 5t2+40t=0 equation: 



5t^2+40t=0

a = 5; b = 40; c = 0;
Δ = b2-4ac
Δ = 402-4·5·0
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-40}{2*5}=\frac{-80}{10}
=-8
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+40}{2*5}=\frac{0}{10}
=0
$

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