20=10t+.5(10)t^2

Solution for 20=10t+.5(10)t^2 equation:

20=10t+.5(10)t^2

We move all terms to the left:

20-(10t+.5(10)t^2)=0

We get rid of parentheses

-.510t^2-10t+20=0

We add all the numbers together, and all the variables

-0.51t^2-10t+20=0

a = -0.51; b = -10; c = +20;
Δ = b2-4ac
Δ = -102-4·(-0.51)·20
Δ = 140.8
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{-(-10)-\sqrt{140.8}}{2*-0.51}=\frac{10-\sqrt{140.8}}{-1.02}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+\sqrt{140.8}}{2*-0.51}=\frac{10+\sqrt{140.8}}{-1.02}
$