2j^2+9=17

Solution for 2j^2+9=17 equation:

2j^2+9=17

We move all terms to the left:

2j^2+9-(17)=0

We add all the numbers together, and all the variables

2j^2-8=0

a = 2; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·2·(-8)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*2}=\frac{-8}{4}
=-2
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*2}=\frac{8}{4}
=2
$