20t-2t^2+3=0

Solution for 20t-2t^2+3=0 equation:

20t-2t^2+3=0

a = -2; b = 20; c = +3;
Δ = b2-4ac
Δ = 202-4·(-2)·3
Δ = 424
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:

$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{106}}{2*-2}=\frac{-20-2\sqrt{106}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{106}}{2*-2}=\frac{-20+2\sqrt{106}}{-4}
$