(t)=4.9t^2+3t=-10

Solution for (t)=4.9t^2+3t=-10 equation:

(t)=4.9t^2+3t=-10

We move all terms to the left:

(t)-(4.9t^2+3t)=0

We get rid of parentheses

-4.9t^2+t-3t=0

We add all the numbers together, and all the variables

-4.9t^2-2t=0

a = -4.9; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-4.9)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-4.9}=\frac{0}{-9.8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-4.9}=\frac{4}{-9.8}
=-4/9.8
$