-16t2+40t+3=15

Solution for -16t2+40t+3=15 equation:

-16t^2+40t+3=15

We move all terms to the left:

-16t^2+40t+3-(15)=0

We add all the numbers together, and all the variables

-16t^2+40t-12=0

a = -16; b = 40; c = -12;
Δ = b2-4ac
Δ = 402-4·(-16)·(-12)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{13}}{2*-16}=\frac{-40-8\sqrt{13}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{13}}{2*-16}=\frac{-40+8\sqrt{13}}{-32}
$