5b^2-4=4

Solution for 5b^2-4=4 equation:

5b^2-4=4

We move all terms to the left:

5b^2-4-(4)=0

We add all the numbers together, and all the variables

5b^2-8=0

a = 5; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·5·(-8)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*5}=\frac{0-4\sqrt{10}}{10}
=-\frac{4\sqrt{10}}{10}
=-\frac{2\sqrt{10}}{5}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*5}=\frac{0+4\sqrt{10}}{10}
=\frac{4\sqrt{10}}{10}
=\frac{2\sqrt{10}}{5}
$