3b^2+8=170

Solution for 3b^2+8=170 equation:

3b^2+8=170

We move all terms to the left:

3b^2+8-(170)=0

We add all the numbers together, and all the variables

3b^2-162=0

a = 3; b = 0; c = -162;
Δ = b2-4ac
Δ = 02-4·3·(-162)
Δ = 1944
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:

$\sqrt{\Delta}=\sqrt{1944}=\sqrt{324*6}=\sqrt{324}*\sqrt{6}=18\sqrt{6}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{6}}{2*3}=\frac{0-18\sqrt{6}}{6}
=-\frac{18\sqrt{6}}{6}
=-3\sqrt{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{6}}{2*3}=\frac{0+18\sqrt{6}}{6}
=\frac{18\sqrt{6}}{6}
=3\sqrt{6}
$