7j^2+8j-2=0

Solution for 7j^2+8j-2=0 equation:

7j^2+8j-2=0

a = 7; b = 8; c = -2;
Δ = b2-4ac
Δ = 82-4·7·(-2)
Δ = 120
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{30}}{2*7}=\frac{-8-2\sqrt{30}}{14}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{30}}{2*7}=\frac{-8+2\sqrt{30}}{14}
$