2j^2+4j-5=0

Solution for 2j^2+4j-5=0 equation:

2j^2+4j-5=0

a = 2; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·2·(-5)
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{14}}{2*2}=\frac{-4-2\sqrt{14}}{4}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{14}}{2*2}=\frac{-4+2\sqrt{14}}{4}
$