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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-3x^2-6x+4x+8)=0

We get rid of parentheses

-3x^2-6x+4x+8=0

We add all the numbers together, and all the variables

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

a = -3; b = -2; c = +8;
Δ = b2-4ac
Δ = -22-4·(-3)·8
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*-3}=\frac{-8}{-6}
=1+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*-3}=\frac{12}{-6}
=-2
$