t^2+33t=0

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Solution for t^2+33t=0 equation: 



t^2+33t=0

a = 1; b = 33; c = 0;
Δ = b2-4ac
Δ = 332-4·1·0
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-33}{2*1}=\frac{-66}{2}
=-33
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+33}{2*1}=\frac{0}{2}
=0
$

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