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-16t^2+16t+280=0
a = -16; b = 16; c = +280;
Δ = b2-4ac
Δ = 162-4·(-16)·280
Δ = 18176
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{18176}=\sqrt{256*71}=\sqrt{256}*\sqrt{71}=16\sqrt{71}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16\sqrt{71}}{2*-16}=\frac{-16-16\sqrt{71}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16\sqrt{71}}{2*-16}=\frac{-16+16\sqrt{71}}{-32} $
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