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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


We calculate terms in parentheses: -(3(x+1)), so:

3(x+1)

We multiply parentheses

3x+3

Back to the equation:

-(3x+3)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-2x-15=0

a = 1; b = -2; c = -15;
Δ = b2-4ac
Δ = -22-4·1·(-15)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*1}=\frac{10}{2}
=5
$