(j+5)(j+5)=20

Solution for (j+5)(j+5)=20 equation:

(j+5)(j+5)=20

We move all terms to the left:

(j+5)(j+5)-(20)=0

We multiply parentheses ..

(+j^2+5j+5j+25)-20=0

We get rid of parentheses

j^2+5j+5j+25-20=0

We add all the numbers together, and all the variables

j^2+10j+5=0

a = 1; b = 10; c = +5;
Δ = b2-4ac
Δ = 102-4·1·5
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{5}}{2*1}=\frac{-10-4\sqrt{5}}{2}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{5}}{2*1}=\frac{-10+4\sqrt{5}}{2}
$