3b^2+16b+19=0

Solution for 3b^2+16b+19=0 equation:

3b^2+16b+19=0

a = 3; b = 16; c = +19;
Δ = b2-4ac
Δ = 162-4·3·19
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{7}}{2*3}=\frac{-16-2\sqrt{7}}{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{7}}{2*3}=\frac{-16+2\sqrt{7}}{6}
$