3b2=316+22

Solution for 3b2=316+22 equation:

3b^2=316+22

We move all terms to the left:

3b^2-(316+22)=0

We add all the numbers together, and all the variables

3b^2-338=0

a = 3; b = 0; c = -338;
Δ = b2-4ac
Δ = 02-4·3·(-338)
Δ = 4056
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:

$\sqrt{\Delta}=\sqrt{4056}=\sqrt{676*6}=\sqrt{676}*\sqrt{6}=26\sqrt{6}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-26\sqrt{6}}{2*3}=\frac{0-26\sqrt{6}}{6}
=-\frac{26\sqrt{6}}{6}
=-\frac{13\sqrt{6}}{3}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+26\sqrt{6}}{2*3}=\frac{0+26\sqrt{6}}{6}
=\frac{26\sqrt{6}}{6}
=\frac{13\sqrt{6}}{3}
$