5b^2-4=21

Solution for 5b^2-4=21 equation:

5b^2-4=21

We move all terms to the left:

5b^2-4-(21)=0

We add all the numbers together, and all the variables

5b^2-25=0

a = 5; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·5·(-25)
Δ = 500
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:

$\sqrt{\Delta}=\sqrt{500}=\sqrt{100*5}=\sqrt{100}*\sqrt{5}=10\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{5}}{2*5}=\frac{0-10\sqrt{5}}{10}
=-\frac{10\sqrt{5}}{10}
=-\sqrt{5}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{5}}{2*5}=\frac{0+10\sqrt{5}}{10}
=\frac{10\sqrt{5}}{10}
=\sqrt{5}
$