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6t^2+7t-3=0
a = 6; b = 7; c = -3;
Δ = b2-4ac
Δ = 72-4·6·(-3)
Δ = 121
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{121}=11$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-11}{2*6}=\frac{-18}{12} =-1+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+11}{2*6}=\frac{4}{12} =1/3 $
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