(x-5)(x+4)=(x-5)(3x+12)

Solution for (x-5)(x+4)=(x-5)(3x+12) equation:

(x-5)(x+4)=(x-5)(3x+12)

We move all terms to the left:

(x-5)(x+4)-((x-5)(3x+12))=0

We multiply parentheses ..

(+x^2+4x-5x-20)-((x-5)(3x+12))=0


We calculate terms in parentheses: -((x-5)(3x+12)), so:

(x-5)(3x+12)

We multiply parentheses ..

(+3x^2+12x-15x-60)

We get rid of parentheses

3x^2+12x-15x-60

We add all the numbers together, and all the variables

3x^2-3x-60

Back to the equation:

-(3x^2-3x-60)


We get rid of parentheses

x^2-3x^2+4x-5x+3x-20+60=0

We add all the numbers together, and all the variables

-2x^2+2x+40=0

a = -2; b = 2; c = +40;
Δ = b2-4ac
Δ = 22-4·(-2)·40
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-18}{2*-2}=\frac{-20}{-4}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+18}{2*-2}=\frac{16}{-4}
=-4
$