4t^2-40t-28=0

Solution for 4t^2-40t-28=0 equation:

4t^2-40t-28=0

a = 4; b = -40; c = -28;
Δ = b2-4ac
Δ = -402-4·4·(-28)
Δ = 2048
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:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-32\sqrt{2}}{2*4}=\frac{40-32\sqrt{2}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+32\sqrt{2}}{2*4}=\frac{40+32\sqrt{2}}{8}
$