t^2+34t=0

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Solution for t^2+34t=0 equation: 



t^2+34t=0

a = 1; b = 34; c = 0;
Δ = b2-4ac
Δ = 342-4·1·0
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-34}{2*1}=\frac{-68}{2}
=-34
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+34}{2*1}=\frac{0}{2}
=0
$

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