-16t^2+112t+4=40

Solution for -16t^2+112t+4=40 equation:

-16t^2+112t+4=40

We move all terms to the left:

-16t^2+112t+4-(40)=0

We add all the numbers together, and all the variables

-16t^2+112t-36=0

a = -16; b = 112; c = -36;
Δ = b2-4ac
Δ = 1122-4·(-16)·(-36)
Δ = 10240
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:

$\sqrt{\Delta}=\sqrt{10240}=\sqrt{1024*10}=\sqrt{1024}*\sqrt{10}=32\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(112)-32\sqrt{10}}{2*-16}=\frac{-112-32\sqrt{10}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(112)+32\sqrt{10}}{2*-16}=\frac{-112+32\sqrt{10}}{-32}
$