(x+4)(x-7)=3(x-4)

Solution for (x+4)(x-7)=3(x-4) equation:

(x+4)(x-7)=3(x-4)

We move all terms to the left:

(x+4)(x-7)-(3(x-4))=0

We multiply parentheses ..

(+x^2-7x+4x-28)-(3(x-4))=0


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

3(x-4)

We multiply parentheses

3x-12

Back to the equation:

-(3x-12)


We get rid of parentheses

x^2-7x+4x-3x-28+12=0

We add all the numbers together, and all the variables

x^2-6x-16=0

a = 1; b = -6; c = -16;
Δ = b2-4ac
Δ = -62-4·1·(-16)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-10}{2*1}=\frac{-4}{2}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+10}{2*1}=\frac{16}{2}
=8
$