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7d^2+41d-6=0
a = 7; b = 41; c = -6;
Δ = b2-4ac
Δ = 412-4·7·(-6)
Δ = 1849
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1849}=43$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-43}{2*7}=\frac{-84}{14} =-6 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+43}{2*7}=\frac{2}{14} =1/7 $
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