2j^2-17j+15=0

Solution for 2j^2-17j+15=0 equation:

2j^2-17j+15=0

a = 2; b = -17; c = +15;
Δ = b2-4ac
Δ = -172-4·2·15
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-13}{2*2}=\frac{4}{4}
=1
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+13}{2*2}=\frac{30}{4}
=7+1/2
$