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16t^2-68t-6=0
a = 16; b = -68; c = -6;
Δ = b2-4ac
Δ = -682-4·16·(-6)
Δ = 5008
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5008}=\sqrt{16*313}=\sqrt{16}*\sqrt{313}=4\sqrt{313}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-68)-4\sqrt{313}}{2*16}=\frac{68-4\sqrt{313}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-68)+4\sqrt{313}}{2*16}=\frac{68+4\sqrt{313}}{32} $
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