16t^2+96t+4=112

Solution for 16t^2+96t+4=112 equation:

16t^2+96t+4=112

We move all terms to the left:

16t^2+96t+4-(112)=0

We add all the numbers together, and all the variables

16t^2+96t-108=0

a = 16; b = 96; c = -108;
Δ = b2-4ac
Δ = 962-4·16·(-108)
Δ = 16128
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:

$\sqrt{\Delta}=\sqrt{16128}=\sqrt{2304*7}=\sqrt{2304}*\sqrt{7}=48\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-48\sqrt{7}}{2*16}=\frac{-96-48\sqrt{7}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+48\sqrt{7}}{2*16}=\frac{-96+48\sqrt{7}}{32}
$