5t^2+24t+8=0

Solution for 5t^2+24t+8=0 equation:

5t^2+24t+8=0

a = 5; b = 24; c = +8;
Δ = b2-4ac
Δ = 242-4·5·8
Δ = 416
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:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{26}}{2*5}=\frac{-24-4\sqrt{26}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{26}}{2*5}=\frac{-24+4\sqrt{26}}{10}
$