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16t^2=800
We move all terms to the left:
16t^2-(800)=0
a = 16; b = 0; c = -800;
Δ = b2-4ac
Δ = 02-4·16·(-800)
Δ = 51200
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{51200}=\sqrt{25600*2}=\sqrt{25600}*\sqrt{2}=160\sqrt{2}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-160\sqrt{2}}{2*16}=\frac{0-160\sqrt{2}}{32} =-\frac{160\sqrt{2}}{32} =-5\sqrt{2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+160\sqrt{2}}{2*16}=\frac{0+160\sqrt{2}}{32} =\frac{160\sqrt{2}}{32} =5\sqrt{2} $
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