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21j^2+14j-56=0
a = 21; b = 14; c = -56;
Δ = b2-4ac
Δ = 142-4·21·(-56)
Δ = 4900
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{4900}=70$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-70}{2*21}=\frac{-84}{42} =-2 $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+70}{2*21}=\frac{56}{42} =1+1/3 $
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