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12t^2+94t-16=0
a = 12; b = 94; c = -16;
Δ = b2-4ac
Δ = 942-4·12·(-16)
Δ = 9604
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9604}=98$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(94)-98}{2*12}=\frac{-192}{24} =-8 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(94)+98}{2*12}=\frac{4}{24} =1/6 $
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