0=-4t^2+40t+3

Solution for 0=-4t^2+40t+3 equation:

0=-4t^2+40t+3

We move all terms to the left:

0-(-4t^2+40t+3)=0

We add all the numbers together, and all the variables

-(-4t^2+40t+3)=0

We get rid of parentheses

4t^2-40t-3=0

a = 4; b = -40; c = -3;
Δ = b2-4ac
Δ = -402-4·4·(-3)
Δ = 1648
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:

$\sqrt{\Delta}=\sqrt{1648}=\sqrt{16*103}=\sqrt{16}*\sqrt{103}=4\sqrt{103}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{103}}{2*4}=\frac{40-4\sqrt{103}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{103}}{2*4}=\frac{40+4\sqrt{103}}{8}
$