(j2-7)/3=25

Solution for (j2-7)/3=25 equation:

(j2-7)/3=25

We move all terms to the left:

(j2-7)/3-(25)=0

We add all the numbers together, and all the variables

(+j^2-7)/3-25=0

We multiply all the terms by the denominator


(+j^2-7)-25*3=0

We add all the numbers together, and all the variables

(+j^2-7)-75=0

We get rid of parentheses

j^2-7-75=0

We add all the numbers together, and all the variables

j^2-82=0

a = 1; b = 0; c = -82;
Δ = b2-4ac
Δ = 02-4·1·(-82)
Δ = 328
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:

$\sqrt{\Delta}=\sqrt{328}=\sqrt{4*82}=\sqrt{4}*\sqrt{82}=2\sqrt{82}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{82}}{2*1}=\frac{0-2\sqrt{82}}{2}
=-\frac{2\sqrt{82}}{2}
=-\sqrt{82}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{82}}{2*1}=\frac{0+2\sqrt{82}}{2}
=\frac{2\sqrt{82}}{2}
=\sqrt{82}
$