15t+10t2=0

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Solution for 15t+10t2=0 equation:



15t+10t^2=0
a = 10; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·10·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*10}=\frac{-30}{20} =-1+1/2 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*10}=\frac{0}{20} =0 $

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