t^2+3t-1500=0

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Solution for t^2+3t-1500=0 equation: 



t^2+3t-1500=0

a = 1; b = 3; c = -1500;
Δ = b2-4ac
Δ = 32-4·1·(-1500)
Δ = 6009
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{6009}}{2*1}=\frac{-3-\sqrt{6009}}{2}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{6009}}{2*1}=\frac{-3+\sqrt{6009}}{2}
$

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