120+10t+-5t2=0

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Solution for 120+10t+-5t2=0 equation:



120+10t+-5t^2=0
We add all the numbers together, and all the variables
-5t^2+10t=0
a = -5; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-5)·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*-5}=\frac{-20}{-10} =+2 $
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-5}=\frac{0}{-10} =0 $

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