5t^2-40t+4=0

Solution for 5t^2-40t+4=0 equation:

5t^2-40t+4=0

a = 5; b = -40; c = +4;
Δ = b2-4ac
Δ = -402-4·5·4
Δ = 1520
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:

$\sqrt{\Delta}=\sqrt{1520}=\sqrt{16*95}=\sqrt{16}*\sqrt{95}=4\sqrt{95}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{95}}{2*5}=\frac{40-4\sqrt{95}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{95}}{2*5}=\frac{40+4\sqrt{95}}{10}
$