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-16t^2+8t+4=0
a = -16; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·(-16)·4
Δ = 320
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{5}}{2*-16}=\frac{-8-8\sqrt{5}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{5}}{2*-16}=\frac{-8+8\sqrt{5}}{-32} $
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