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j-9/5j=4
We move all terms to the left:
j-9/5j-(4)=0
Domain of the equation: 5j!=0We multiply all the terms by the denominator
j!=0/5
j!=0
j∈R
j*5j-4*5j-9=0
Wy multiply elements
5j^2-20j-9=0
a = 5; b = -20; c = -9;
Δ = b2-4ac
Δ = -202-4·5·(-9)
Δ = 580
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}$$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{580}=\sqrt{4*145}=\sqrt{4}*\sqrt{145}=2\sqrt{145}$$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{145}}{2*5}=\frac{20-2\sqrt{145}}{10} $$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{145}}{2*5}=\frac{20+2\sqrt{145}}{10} $
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