5t^2-20t+6=0

Solution for 5t^2-20t+6=0 equation:

5t^2-20t+6=0

a = 5; b = -20; c = +6;
Δ = b2-4ac
Δ = -202-4·5·6
Δ = 280
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:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{70}}{2*5}=\frac{20-2\sqrt{70}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{70}}{2*5}=\frac{20+2\sqrt{70}}{10}
$