3b^2+7=28

Solution for 3b^2+7=28 equation:

3b^2+7=28

We move all terms to the left:

3b^2+7-(28)=0

We add all the numbers together, and all the variables

3b^2-21=0

a = 3; b = 0; c = -21;
Δ = b2-4ac
Δ = 02-4·3·(-21)
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{7}}{2*3}=\frac{0-6\sqrt{7}}{6}
=-\frac{6\sqrt{7}}{6}
=-\sqrt{7}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{7}}{2*3}=\frac{0+6\sqrt{7}}{6}
=\frac{6\sqrt{7}}{6}
=\sqrt{7}
$