2t^2+3t-24=0

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Solution for 2t^2+3t-24=0 equation: 



2t^2+3t-24=0

a = 2; b = 3; c = -24;
Δ = b2-4ac
Δ = 32-4·2·(-24)
Δ = 201
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{201}}{2*2}=\frac{-3-\sqrt{201}}{4}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{201}}{2*2}=\frac{-3+\sqrt{201}}{4}
$

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