-2(j-11+3)j+4=8

Solution for -2(j-11+3)j+4=8 equation:

-2(j-11+3)j+4=8

We move all terms to the left:

-2(j-11+3)j+4-(8)=0

We add all the numbers together, and all the variables

-2(j-8)j+4-8=0

We add all the numbers together, and all the variables

-2(j-8)j-4=0

We multiply parentheses

-2j^2+16j-4=0

a = -2; b = 16; c = -4;
Δ = b2-4ac
Δ = 162-4·(-2)·(-4)
Δ = 224
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:

$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{14}}{2*-2}=\frac{-16-4\sqrt{14}}{-4}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{14}}{2*-2}=\frac{-16+4\sqrt{14}}{-4}
$