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6x^2+41x-7=0
a = 6; b = 41; c = -7;
Δ = b2-4ac
Δ = 412-4·6·(-7)
Δ = 1849
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{1849}=43$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-43}{2*6}=\frac{-84}{12} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+43}{2*6}=\frac{2}{12} =1/6 $
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