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42x^2-43x+6=0
a = 42; b = -43; c = +6;
Δ = b2-4ac
Δ = -432-4·42·6
Δ = 841
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{841}=29$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-43)-29}{2*42}=\frac{14}{84} =1/6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+29}{2*42}=\frac{72}{84} =6/7 $
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