50(t)=-16t^2+63t+4

Solution for 50(t)=-16t^2+63t+4 equation:

50(t)=-16t^2+63t+4

We move all terms to the left:

50(t)-(-16t^2+63t+4)=0

We get rid of parentheses

16t^2-63t+50t-4=0

We add all the numbers together, and all the variables

16t^2-13t-4=0

a = 16; b = -13; c = -4;
Δ = b2-4ac
Δ = -132-4·16·(-4)
Δ = 425
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5\sqrt{17}}{2*16}=\frac{13-5\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5\sqrt{17}}{2*16}=\frac{13+5\sqrt{17}}{32}
$