(x+4)(3x+2)=20

Solution for (x+4)(3x+2)=20 equation:

(x+4)(3x+2)=20

We move all terms to the left:

(x+4)(3x+2)-(20)=0

We multiply parentheses ..

(+3x^2+2x+12x+8)-20=0

We get rid of parentheses

3x^2+2x+12x+8-20=0

We add all the numbers together, and all the variables

3x^2+14x-12=0

a = 3; b = 14; c = -12;
Δ = b2-4ac
Δ = 142-4·3·(-12)
Δ = 340
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:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{85}}{2*3}=\frac{-14-2\sqrt{85}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{85}}{2*3}=\frac{-14+2\sqrt{85}}{6}
$