-2j(j+5)+6=4(j+2)

Solution for -2j(j+5)+6=4(j+2) equation:

-2j(j+5)+6=4(j+2)

We move all terms to the left:

-2j(j+5)+6-(4(j+2))=0

We multiply parentheses

-2j^2-10j-(4(j+2))+6=0


We calculate terms in parentheses: -(4(j+2)), so:

4(j+2)

We multiply parentheses

4j+8

Back to the equation:

-(4j+8)


We get rid of parentheses

-2j^2-10j-4j-8+6=0

We add all the numbers together, and all the variables

-2j^2-14j-2=0

a = -2; b = -14; c = -2;
Δ = b2-4ac
Δ = -142-4·(-2)·(-2)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6\sqrt{5}}{2*-2}=\frac{14-6\sqrt{5}}{-4}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6\sqrt{5}}{2*-2}=\frac{14+6\sqrt{5}}{-4}
$