40t^2-50t+4=0

Solution for 40t^2-50t+4=0 equation:

40t^2-50t+4=0

a = 40; b = -50; c = +4;
Δ = b2-4ac
Δ = -502-4·40·4
Δ = 1860
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:

$\sqrt{\Delta}=\sqrt{1860}=\sqrt{4*465}=\sqrt{4}*\sqrt{465}=2\sqrt{465}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{465}}{2*40}=\frac{50-2\sqrt{465}}{80}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{465}}{2*40}=\frac{50+2\sqrt{465}}{80}
$