39=11t+(4.9t^2)

Solution for 39=11t+(4.9t^2) equation:

39=11t+(4.9t^2)

We move all terms to the left:

39-(11t+(4.9t^2))=0


We calculate terms in parentheses: -(11t+(4.9t^2)), so:

11t+(4.9t^2)

We get rid of parentheses

4.9t^2+11t

Back to the equation:

-(4.9t^2+11t)


We get rid of parentheses

-4.9t^2-11t+39=0

a = -4.9; b = -11; c = +39;
Δ = b2-4ac
Δ = -112-4·(-4.9)·39
Δ = 885.4
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{-(-11)-\sqrt{885.4}}{2*-4.9}=\frac{11-\sqrt{885.4}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{885.4}}{2*-4.9}=\frac{11+\sqrt{885.4}}{-9.8}
$