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7j^2+60j-27=0
a = 7; b = 60; c = -27;
Δ = b2-4ac
Δ = 602-4·7·(-27)
Δ = 4356
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{4356}=66$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-66}{2*7}=\frac{-126}{14} =-9 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+66}{2*7}=\frac{6}{14} =3/7 $
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