5j^2=3j+4

Solution for 5j^2=3j+4 equation:

5j^2=3j+4

We move all terms to the left:

5j^2-(3j+4)=0

We get rid of parentheses

5j^2-3j-4=0

a = 5; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·5·(-4)
Δ = 89
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{89}}{2*5}=\frac{3-\sqrt{89}}{10}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{89}}{2*5}=\frac{3+\sqrt{89}}{10}
$