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7j^2-35j-168=0
a = 7; b = -35; c = -168;
Δ = b2-4ac
Δ = -352-4·7·(-168)
Δ = 5929
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{5929}=77$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-77}{2*7}=\frac{-42}{14} =-3 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+77}{2*7}=\frac{112}{14} =8 $
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