3b^2+16b+19=5b^2

Solution for 3b^2+16b+19=5b^2 equation:

3b^2+16b+19=5b^2

We move all terms to the left:

3b^2+16b+19-(5b^2)=0

determiningTheFunctionDomain
3b^2-5b^2+16b+19=0

We add all the numbers together, and all the variables

-2b^2+16b+19=0

a = -2; b = 16; c = +19;
Δ = b2-4ac
Δ = 162-4·(-2)·19
Δ = 408
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:

$\sqrt{\Delta}=\sqrt{408}=\sqrt{4*102}=\sqrt{4}*\sqrt{102}=2\sqrt{102}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{102}}{2*-2}=\frac{-16-2\sqrt{102}}{-4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{102}}{2*-2}=\frac{-16+2\sqrt{102}}{-4}
$