J(8j+5)=8j^2+5

Solution for J(8j+5)=8j^2+5 equation:

(8J+5)=8J^2+5

We move all terms to the left:

(8J+5)-(8J^2+5)=0

We get rid of parentheses

-8J^2+8J+5-5=0

We add all the numbers together, and all the variables

-8J^2+8J=0

a = -8; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·(-8)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$J_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$J_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$J_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*-8}=\frac{-16}{-16}
=1
$
$J_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*-8}=\frac{0}{-16}
=0
$