t^2+3t-30=0

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Solution for t^2+3t-30=0 equation: 



t^2+3t-30=0

a = 1; b = 3; c = -30;
Δ = b2-4ac
Δ = 32-4·1·(-30)
Δ = 129
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{129}}{2*1}=\frac{-3-\sqrt{129}}{2}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{129}}{2*1}=\frac{-3+\sqrt{129}}{2}
$

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