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

Solution for 0=(x+3)(x-4) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+x^2-4x+3x-12)), so:

(+x^2-4x+3x-12)

We get rid of parentheses

x^2-4x+3x-12

We add all the numbers together, and all the variables

x^2-1x-12

Back to the equation:

-(x^2-1x-12)


We get rid of parentheses

-x^2+1x+12=0

We add all the numbers together, and all the variables

-1x^2+x+12=0

a = -1; b = 1; c = +12;
Δ = b2-4ac
Δ = 12-4·(-1)·12
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-7}{2*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+7}{2*-1}=\frac{6}{-2}
=-3
$