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-16t^2+248t+8=0
a = -16; b = 248; c = +8;
Δ = b2-4ac
Δ = 2482-4·(-16)·8
Δ = 62016
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{62016}=\sqrt{64*969}=\sqrt{64}*\sqrt{969}=8\sqrt{969}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(248)-8\sqrt{969}}{2*-16}=\frac{-248-8\sqrt{969}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(248)+8\sqrt{969}}{2*-16}=\frac{-248+8\sqrt{969}}{-32} $
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