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-16t^2-6t+24=0
a = -16; b = -6; c = +24;
Δ = b2-4ac
Δ = -62-4·(-16)·24
Δ = 1572
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1572}=\sqrt{4*393}=\sqrt{4}*\sqrt{393}=2\sqrt{393}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{393}}{2*-16}=\frac{6-2\sqrt{393}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{393}}{2*-16}=\frac{6+2\sqrt{393}}{-32} $
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