(7j+9)(j-3)=0

Solution for (7j+9)(j-3)=0 equation:

(7j+9)(j-3)=0

We multiply parentheses ..

(+7j^2-21j+9j-27)=0

We get rid of parentheses

7j^2-21j+9j-27=0

We add all the numbers together, and all the variables

7j^2-12j-27=0

a = 7; b = -12; c = -27;
Δ = b2-4ac
Δ = -122-4·7·(-27)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-30}{2*7}=\frac{-18}{14}
=-1+2/7
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+30}{2*7}=\frac{42}{14}
=3
$