t=4.9t^2+20t+10

Solution for t=4.9t^2+20t+10 equation:

t=4.9t^2+20t+10

We move all terms to the left:

t-(4.9t^2+20t+10)=0

We get rid of parentheses

-4.9t^2+t-20t-10=0

We add all the numbers together, and all the variables

-4.9t^2-19t-10=0

a = -4.9; b = -19; c = -10;
Δ = b2-4ac
Δ = -192-4·(-4.9)·(-10)
Δ = 165
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{-(-19)-\sqrt{165}}{2*-4.9}=\frac{19-\sqrt{165}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{165}}{2*-4.9}=\frac{19+\sqrt{165}}{-9.8}
$