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16t^2-106t-25=0
a = 16; b = -106; c = -25;
Δ = b2-4ac
Δ = -1062-4·16·(-25)
Δ = 12836
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12836}=\sqrt{4*3209}=\sqrt{4}*\sqrt{3209}=2\sqrt{3209}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-106)-2\sqrt{3209}}{2*16}=\frac{106-2\sqrt{3209}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-106)+2\sqrt{3209}}{2*16}=\frac{106+2\sqrt{3209}}{32} $
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