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-16t^2+68t+15=0
a = -16; b = 68; c = +15;
Δ = b2-4ac
Δ = 682-4·(-16)·15
Δ = 5584
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5584}=\sqrt{16*349}=\sqrt{16}*\sqrt{349}=4\sqrt{349}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-4\sqrt{349}}{2*-16}=\frac{-68-4\sqrt{349}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+4\sqrt{349}}{2*-16}=\frac{-68+4\sqrt{349}}{-32} $
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