(4t^2)+24t-50=0

Solution for (4t^2)+24t-50=0 equation:

(4t^2)+24t-50=0

a = 4; b = 24; c = -50;
Δ = b2-4ac
Δ = 242-4·4·(-50)
Δ = 1376
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1376}=\sqrt{16*86}=\sqrt{16}*\sqrt{86}=4\sqrt{86}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{86}}{2*4}=\frac{-24-4\sqrt{86}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{86}}{2*4}=\frac{-24+4\sqrt{86}}{8}
$