3x(x-4)=4x-18

Solution for 3x(x-4)=4x-18 equation:

3x(x-4)=4x-18

We move all terms to the left:

3x(x-4)-(4x-18)=0

We multiply parentheses

3x^2-12x-(4x-18)=0

We get rid of parentheses

3x^2-12x-4x+18=0

We add all the numbers together, and all the variables

3x^2-16x+18=0

a = 3; b = -16; c = +18;
Δ = b2-4ac
Δ = -162-4·3·18
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{10}}{2*3}=\frac{16-2\sqrt{10}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{10}}{2*3}=\frac{16+2\sqrt{10}}{6}
$