7j^2+29j+4=0

Solution for 7j^2+29j+4=0 equation:

7j^2+29j+4=0

a = 7; b = 29; c = +4;
Δ = b2-4ac
Δ = 292-4·7·4
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-27}{2*7}=\frac{-56}{14}
=-4
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+27}{2*7}=\frac{-2}{14}
=-1/7
$