X=10+20t-5t2

Solution for X=10+20t-5t2 equation:

X=10+20X-5X^2

We move all terms to the left:

X-(10+20X-5X^2)=0

We get rid of parentheses

5X^2-20X+X-10=0

We add all the numbers together, and all the variables

5X^2-19X-10=0

a = 5; b = -19; c = -10;
Δ = b2-4ac
Δ = -192-4·5·(-10)
Δ = 561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{561}}{2*5}=\frac{19-\sqrt{561}}{10}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{561}}{2*5}=\frac{19+\sqrt{561}}{10}
$