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-32t^2+64t+6=0
a = -32; b = 64; c = +6;
Δ = b2-4ac
Δ = 642-4·(-32)·6
Δ = 4864
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4864}=\sqrt{256*19}=\sqrt{256}*\sqrt{19}=16\sqrt{19}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-16\sqrt{19}}{2*-32}=\frac{-64-16\sqrt{19}}{-64} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+16\sqrt{19}}{2*-32}=\frac{-64+16\sqrt{19}}{-64} $
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