(X-3)(4x+1)+2(3-x)(x+2)=0

Solution for (X-3)(4x+1)+2(3-x)(x+2)=0 equation:

(X-3)(4X+1)+2(3-X)(X+2)=0

We add all the numbers together, and all the variables

(X-3)(4X+1)+2(-1X+3)(X+2)=0

We multiply parentheses ..

(+4X^2+X-12X-3)+2(-1X+3)(X+2)=0

We get rid of parentheses

4X^2+X-12X+2(-1X+3)(X+2)-3=0

We multiply parentheses ..

4X^2+2(-1X^2-2X+3X+6)+X-12X-3=0

We add all the numbers together, and all the variables

4X^2+2(-1X^2-2X+3X+6)-11X-3=0

We multiply parentheses

4X^2-2X^2-4X+6X-11X+12-3=0

We add all the numbers together, and all the variables

2X^2-9X+9=0

a = 2; b = -9; c = +9;
Δ = b2-4ac
Δ = -92-4·2·9
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3}{2*2}=\frac{6}{4}
=1+1/2
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3}{2*2}=\frac{12}{4}
=3
$