3x^2+47X+132=0

Solution for 3x^2+47X+132=0 equation:

3x^2+47x+132=0

a = 3; b = 47; c = +132;
Δ = b2-4ac
Δ = 472-4·3·132
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-25}{2*3}=\frac{-72}{6}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+25}{2*3}=\frac{-22}{6}
=-3+2/3
$