x(3x+4)=12

Solution for x(3x+4)=12 equation:

x(3x+4)=12

We move all terms to the left:

x(3x+4)-(12)=0

We multiply parentheses

3x^2+4x-12=0

a = 3; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·3·(-12)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{10}}{2*3}=\frac{-4-4\sqrt{10}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{10}}{2*3}=\frac{-4+4\sqrt{10}}{6}
$