-7=j2+-8

Solution for -7=j2+-8 equation:

-7=j2+-8

We move all terms to the left:

-7-(j2+-8)=0

We add all the numbers together, and all the variables

-(+j^2+-8)-7=0

We use the square of the difference formula

-(+j^2-8)-7=0

We get rid of parentheses

-j^2+8-7=0

We add all the numbers together, and all the variables

-1j^2+1=0

a = -1; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-1)·1
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2}{2*-1}=\frac{-2}{-2}
=1
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2}{2*-1}=\frac{2}{-2}
=-1
$