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-75=-16t^2
We move all terms to the left:
-75-(-16t^2)=0
We get rid of parentheses
16t^2-75=0
a = 16; b = 0; c = -75;
Δ = b2-4ac
Δ = 02-4·16·(-75)
Δ = 4800
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40\sqrt{3}}{2*16}=\frac{0-40\sqrt{3}}{32} =-\frac{40\sqrt{3}}{32} =-\frac{5\sqrt{3}}{4} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{3}}{2*16}=\frac{0+40\sqrt{3}}{32} =\frac{40\sqrt{3}}{32} =\frac{5\sqrt{3}}{4} $
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