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50=16t^2
We move all terms to the left:
50-(16t^2)=0
a = -16; b = 0; c = +50;
Δ = b2-4ac
Δ = 02-4·(-16)·50
Δ = 3200
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{2}}{2*-16}=\frac{0-40\sqrt{2}}{-32} =-\frac{40\sqrt{2}}{-32} =-\frac{5\sqrt{2}}{-4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{2}}{2*-16}=\frac{0+40\sqrt{2}}{-32} =\frac{40\sqrt{2}}{-32} =\frac{5\sqrt{2}}{-4} $
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