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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(x-1)

We get rid of parentheses

x-1

Back to the equation:

-(x-1)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-2x-11=0

a = 1; b = -2; c = -11;
Δ = b2-4ac
Δ = -22-4·1·(-11)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4\sqrt{3}}{2*1}=\frac{2-4\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{3}}{2*1}=\frac{2+4\sqrt{3}}{2}
$