4t^2+10t=0

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Solution for 4t^2+10t=0 equation: 



4t^2+10t=0

a = 4; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·4·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*4}=\frac{-20}{8}
=-2+1/2
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*4}=\frac{0}{8}
=0
$

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