4t2-20t=0

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Solution for 4t2-20t=0 equation:



4t^2-20t=0
a = 4; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·4·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*4}=\frac{0}{8} =0 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*4}=\frac{40}{8} =5 $

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