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7j^2+50j+7=0
a = 7; b = 50; c = +7;
Δ = b2-4ac
Δ = 502-4·7·7
Δ = 2304
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{2304}=48$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-48}{2*7}=\frac{-98}{14} =-7 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+48}{2*7}=\frac{-2}{14} =-1/7 $
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