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-16t^2+32t+2=0
a = -16; b = 32; c = +2;
Δ = b2-4ac
Δ = 322-4·(-16)·2
Δ = 1152
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1152}=\sqrt{576*2}=\sqrt{576}*\sqrt{2}=24\sqrt{2}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-24\sqrt{2}}{2*-16}=\frac{-32-24\sqrt{2}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+24\sqrt{2}}{2*-16}=\frac{-32+24\sqrt{2}}{-32} $
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