12j^2+16j-28=0

Solution for 12j^2+16j-28=0 equation:

12j^2+16j-28=0

a = 12; b = 16; c = -28;
Δ = b2-4ac
Δ = 162-4·12·(-28)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-40}{2*12}=\frac{-56}{24}
=-2+1/3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+40}{2*12}=\frac{24}{24}
=1
$