18j^2+96j+78=0

Solution for 18j^2+96j+78=0 equation:

18j^2+96j+78=0

a = 18; b = 96; c = +78;
Δ = b2-4ac
Δ = 962-4·18·78
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-60}{2*18}=\frac{-156}{36}
=-4+1/3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+60}{2*18}=\frac{-36}{36}
=-1
$