17t2+7t=0

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Solution for 17t2+7t=0 equation: 



17t^2+7t=0

a = 17; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·17·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*17}=\frac{-14}{34}
=-7/17
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*17}=\frac{0}{34}
=0
$

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