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6x^2-3-7x=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:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-11}{2*6}=\frac{-4}{12} =-1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+11}{2*6}=\frac{18}{12} =1+1/2 $
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