2j^2+17j+8=0

Solution for 2j^2+17j+8=0 equation:

2j^2+17j+8=0

a = 2; b = 17; c = +8;
Δ = b2-4ac
Δ = 172-4·2·8
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-15}{2*2}=\frac{-32}{4}
=-8
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+15}{2*2}=\frac{-2}{4}
=-1/2
$