J(x)=-4/5x+7J(x)=7

Solution for J(x)=-4/5x+7J(x)=7 equation:

(J)=-4/5J+7(J)=7

We move all terms to the left:

(J)-(-4/5J+7(J))=0


Domain of the equation: 5J+7J)!=0

J∈R


We add all the numbers together, and all the variables

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}
$