7j(j+2)=-2j+12=

Solution for 7j(j+2)=-2j+12= equation:

7j(j+2)=-2j+12=

We move all terms to the left:

7j(j+2)-(-2j+12)=0

We multiply parentheses

7j^2+14j-(-2j+12)=0

We get rid of parentheses

7j^2+14j+2j-12=0

We add all the numbers together, and all the variables

7j^2+16j-12=0

a = 7; b = 16; c = -12;
Δ = b2-4ac
Δ = 162-4·7·(-12)
Δ = 592
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:

$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{37}}{2*7}=\frac{-16-4\sqrt{37}}{14}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{37}}{2*7}=\frac{-16+4\sqrt{37}}{14}
$