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3b^2+7b-6=0
a = 3; b = 7; c = -6;
Δ = b2-4ac
Δ = 72-4·3·(-6)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-11}{2*3}=\frac{-18}{6} =-3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+11}{2*3}=\frac{4}{6} =2/3 $
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