20=1.5(2x+4)(x+3)

Solution for 20=1.5(2x+4)(x+3) equation:

20=1.5(2x+4)(x+3)

We move all terms to the left:

20-(1.5(2x+4)(x+3))=0

We multiply parentheses ..

-(1.5(+2x^2+6x+4x+12))+20=0


We calculate terms in parentheses: -(1.5(+2x^2+6x+4x+12)), so:

1.5(+2x^2+6x+4x+12)

We multiply parentheses

3x^2+9x+6x+18

We add all the numbers together, and all the variables

3x^2+15x+18

Back to the equation:

-(3x^2+15x+18)


We get rid of parentheses

-3x^2-15x-18+20=0

We add all the numbers together, and all the variables

-3x^2-15x+2=0

a = -3; b = -15; c = +2;
Δ = b2-4ac
Δ = -152-4·(-3)·2
Δ = 249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{249}}{2*-3}=\frac{15-\sqrt{249}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{249}}{2*-3}=\frac{15+\sqrt{249}}{-6}
$