3x+4(x-5)=3x(x-4)

Solution for 3x+4(x-5)=3x(x-4) equation:

3x+4(x-5)=3x(x-4)

We move all terms to the left:

3x+4(x-5)-(3x(x-4))=0

We multiply parentheses

3x+4x-(3x(x-4))-20=0


We calculate terms in parentheses: -(3x(x-4)), so:

3x(x-4)

We multiply parentheses

3x^2-12x

Back to the equation:

-(3x^2-12x)


We add all the numbers together, and all the variables

7x-(3x^2-12x)-20=0

We get rid of parentheses

-3x^2+7x+12x-20=0

We add all the numbers together, and all the variables

-3x^2+19x-20=0

a = -3; b = 19; c = -20;
Δ = b2-4ac
Δ = 192-4·(-3)·(-20)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-11}{2*-3}=\frac{-30}{-6}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+11}{2*-3}=\frac{-8}{-6}
=1+1/3
$