-16t^2+40t=4

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

-16t^2+40t=4

We move all terms to the left:

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

a = -16; b = 40; c = -4;
Δ = b2-4ac
Δ = 402-4·(-16)·(-4)
Δ = 1344
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:

$\sqrt{\Delta}=\sqrt{1344}=\sqrt{64*21}=\sqrt{64}*\sqrt{21}=8\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{21}}{2*-16}=\frac{-40-8\sqrt{21}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{21}}{2*-16}=\frac{-40+8\sqrt{21}}{-32}
$