4t^2-24t-16=0

Solution for 4t^2-24t-16=0 equation:

4t^2-24t-16=0

a = 4; b = -24; c = -16;
Δ = b2-4ac
Δ = -242-4·4·(-16)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-8\sqrt{13}}{2*4}=\frac{24-8\sqrt{13}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+8\sqrt{13}}{2*4}=\frac{24+8\sqrt{13}}{8}
$