50=2t+4t^2

Solution for 50=2t+4t^2 equation:

50=2t+4t^2

We move all terms to the left:

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

We get rid of parentheses

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

a = -4; b = -2; c = +50;
Δ = b2-4ac
Δ = -22-4·(-4)·50
Δ = 804
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:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{201}}{2*-4}=\frac{2-2\sqrt{201}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{201}}{2*-4}=\frac{2+2\sqrt{201}}{-8}
$