58j=-28-4j^2

Solution for 58j=-28-4j^2 equation:

58j=-28-4j^2

We move all terms to the left:

58j-(-28-4j^2)=0

We get rid of parentheses

4j^2+58j+28=0

a = 4; b = 58; c = +28;
Δ = b2-4ac
Δ = 582-4·4·28
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(58)-54}{2*4}=\frac{-112}{8}
=-14
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(58)+54}{2*4}=\frac{-4}{8}
=-1/2
$