4j^2=8j+3

Solution for 4j^2=8j+3 equation:

4j^2=8j+3

We move all terms to the left:

4j^2-(8j+3)=0

We get rid of parentheses

4j^2-8j-3=0

a = 4; b = -8; c = -3;
Δ = b2-4ac
Δ = -82-4·4·(-3)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{7}}{2*4}=\frac{8-4\sqrt{7}}{8}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{7}}{2*4}=\frac{8+4\sqrt{7}}{8}
$