18j^2-240j+78=0

Solution for 18j^2-240j+78=0 equation:

18j^2-240j+78=0

a = 18; b = -240; c = +78;
Δ = b2-4ac
Δ = -2402-4·18·78
Δ = 51984
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{51984}=228$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-228}{2*18}=\frac{12}{36}
=1/3
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+228}{2*18}=\frac{468}{36}
=13
$