4j^2+96j+92=0

Solution for 4j^2+96j+92=0 equation:

4j^2+96j+92=0

a = 4; b = 96; c = +92;
Δ = b2-4ac
Δ = 962-4·4·92
Δ = 7744
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7744}=88$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-88}{2*4}=\frac{-184}{8}
=-23
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+88}{2*4}=\frac{-8}{8}
=-1
$