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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

-4x^2+12x-2x+6=0

We add all the numbers together, and all the variables

-4x^2+10x+6=0

a = -4; b = 10; c = +6;
Δ = b2-4ac
Δ = 102-4·(-4)·6
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14}{2*-4}=\frac{-24}{-8}
=+3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*-4}=\frac{4}{-8}
=-1/2
$