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-16t^2+20t+20=0
a = -16; b = 20; c = +20;
Δ = b2-4ac
Δ = 202-4·(-16)·20
Δ = 1680
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1680}=\sqrt{16*105}=\sqrt{16}*\sqrt{105}=4\sqrt{105}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{105}}{2*-16}=\frac{-20-4\sqrt{105}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{105}}{2*-16}=\frac{-20+4\sqrt{105}}{-32} $
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