18b^2+150b+168=0

Solution for 18b^2+150b+168=0 equation:

18b^2+150b+168=0

a = 18; b = 150; c = +168;
Δ = b2-4ac
Δ = 1502-4·18·168
Δ = 10404
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10404}=102$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(150)-102}{2*18}=\frac{-252}{36}
=-7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(150)+102}{2*18}=\frac{-48}{36}
=-1+1/3
$