4t^2+5t-24=0

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Solution for 4t^2+5t-24=0 equation: 



4t^2+5t-24=0

a = 4; b = 5; c = -24;
Δ = b2-4ac
Δ = 52-4·4·(-24)
Δ = 409
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{-(5)-\sqrt{409}}{2*4}=\frac{-5-\sqrt{409}}{8}
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{409}}{2*4}=\frac{-5+\sqrt{409}}{8}
$

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