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

Solution for 2=(x+3)(x-4) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

-1x^2+x+14=0

a = -1; b = 1; c = +14;
Δ = b2-4ac
Δ = 12-4·(-1)·14
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{-(1)-\sqrt{57}}{2*-1}=\frac{-1-\sqrt{57}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{57}}{2*-1}=\frac{-1+\sqrt{57}}{-2}
$