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

Solution for (3x+4)(x+2)-6x^2=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2+10x+8=0

a = -3; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·(-3)·8
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*-3}=\frac{-24}{-6}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*-3}=\frac{4}{-6}
=-2/3
$