4j^2+43j-11=0

Solution for 4j^2+43j-11=0 equation:

4j^2+43j-11=0

a = 4; b = 43; c = -11;
Δ = b2-4ac
Δ = 432-4·4·(-11)
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-45}{2*4}=\frac{-88}{8}
=-11
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+45}{2*4}=\frac{2}{8}
=1/4
$