16t^2+92t+24=0

Solution for 16t^2+92t+24=0 equation:

16t^2+92t+24=0

a = 16; b = 92; c = +24;
Δ = b2-4ac
Δ = 922-4·16·24
Δ = 6928
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:

$\sqrt{\Delta}=\sqrt{6928}=\sqrt{16*433}=\sqrt{16}*\sqrt{433}=4\sqrt{433}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(92)-4\sqrt{433}}{2*16}=\frac{-92-4\sqrt{433}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(92)+4\sqrt{433}}{2*16}=\frac{-92+4\sqrt{433}}{32}
$