0=78.4t+4.9(t)^2

Solution for 0=78.4t+4.9(t)^2 equation:

0=78.4t+4.9(t)^2

We move all terms to the left:

0-(78.4t+4.9(t)^2)=0

We add all the numbers together, and all the variables

-(78.4t+4.9t^2)=0

We get rid of parentheses

-4.9t^2-78.4t=0

a = -4.9; b = -78.4; c = 0;
Δ = b2-4ac
Δ = -78.42-4·(-4.9)·0
Δ = 6146.56
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{-(-78.4)-\sqrt{6146.56}}{2*-4.9}=\frac{78.4-\sqrt{6146.56}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-78.4)+\sqrt{6146.56}}{2*-4.9}=\frac{78.4+\sqrt{6146.56}}{-9.8}
$