20t^2-3t=20

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

20t^2-3t=20

We move all terms to the left:

20t^2-3t-(20)=0

a = 20; b = -3; c = -20;
Δ = b2-4ac
Δ = -32-4·20·(-20)
Δ = 1609
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{1609}}{2*20}=\frac{3-\sqrt{1609}}{40}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{1609}}{2*20}=\frac{3+\sqrt{1609}}{40}
$