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(J)=-4/5J+7(J)=-5
We move all terms to the left:
(J)-(-4/5J+7(J))=0
Domain of the equation: 5J+7J)!=0We add all the numbers together, and all the variables
J∈R
J-(+7J-4/5J)=0
We get rid of parentheses
J-7J+4/5J=0
We multiply all the terms by the denominator
J*5J-7J*5J+4=0
Wy multiply elements
5J^2-35J^2+4=0
We add all the numbers together, and all the variables
-30J^2+4=0
a = -30; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-30)·4
Δ = 480
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}$$J_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$$J_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{30}}{2*-30}=\frac{0-4\sqrt{30}}{-60} =-\frac{4\sqrt{30}}{-60} =-\frac{\sqrt{30}}{-15} $$J_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{30}}{2*-30}=\frac{0+4\sqrt{30}}{-60} =\frac{4\sqrt{30}}{-60} =\frac{\sqrt{30}}{-15} $
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