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

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

(J)=-4/5J+7

We move all terms to the left:

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


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

J∈R


We get rid of parentheses

J+4/5J-7=0

We multiply all the terms by the denominator


J*5J-7*5J+4=0

Wy multiply elements

5J^2-35J+4=0

a = 5; b = -35; c = +4;
Δ = b2-4ac
Δ = -352-4·5·4
Δ = 1145
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$J_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$J_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$J_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-\sqrt{1145}}{2*5}=\frac{35-\sqrt{1145}}{10}
$
$J_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{1145}}{2*5}=\frac{35+\sqrt{1145}}{10}
$