(x+3)(x-4)=4(x-1)

Solution for (x+3)(x-4)=4(x-1) equation:

(x+3)(x-4)=4(x-1)

We move all terms to the left:

(x+3)(x-4)-(4(x-1))=0

We multiply parentheses ..

(+x^2-4x+3x-12)-(4(x-1))=0


We calculate terms in parentheses: -(4(x-1)), so:

4(x-1)

We multiply parentheses

4x-4

Back to the equation:

-(4x-4)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-5x-8=0

a = 1; b = -5; c = -8;
Δ = b2-4ac
Δ = -52-4·1·(-8)
Δ = 57
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{57}}{2*1}=\frac{5-\sqrt{57}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{57}}{2*1}=\frac{5+\sqrt{57}}{2}
$