2x(3x+4)=32

Solution for 2x(3x+4)=32 equation:

2x(3x+4)=32

We move all terms to the left:

2x(3x+4)-(32)=0

We multiply parentheses

6x^2+8x-32=0

a = 6; b = 8; c = -32;
Δ = b2-4ac
Δ = 82-4·6·(-32)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{13}}{2*6}=\frac{-8-8\sqrt{13}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{13}}{2*6}=\frac{-8+8\sqrt{13}}{12}
$