28(t)=-16t^2=50t

Solution for 28(t)=-16t^2=50t equation:

28(t)=-16t^2=50t

We move all terms to the left:

28(t)-(-16t^2)=0

We get rid of parentheses

16t^2+28t=0

a = 16; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·16·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*16}=\frac{-56}{32}
=-1+3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*16}=\frac{0}{32}
=0
$