4+(3c)2=20

Solution for 4+(3c)2=20 equation:

4+(3c)2=20

We move all terms to the left:

4+(3c)2-(20)=0

We add all the numbers together, and all the variables

3c^2-16=0

a = 3; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·3·(-16)
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*3}=\frac{0-8\sqrt{3}}{6}
=-\frac{8\sqrt{3}}{6}
=-\frac{4\sqrt{3}}{3}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*3}=\frac{0+8\sqrt{3}}{6}
=\frac{8\sqrt{3}}{6}
=\frac{4\sqrt{3}}{3}
$