8t^2+32t=0

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Solution for 8t^2+32t=0 equation: 



8t^2+32t=0

a = 8; b = 32; c = 0;
Δ = b2-4ac
Δ = 322-4·8·0
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-32}{2*8}=\frac{-64}{16}
=-4
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+32}{2*8}=\frac{0}{16}
=0
$

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