(x+4)(x-5)=3-(x+7)

Solution for (x+4)(x-5)=3-(x+7) equation:

(x+4)(x-5)=3-(x+7)

We move all terms to the left:

(x+4)(x-5)-(3-(x+7))=0

We multiply parentheses ..

(+x^2-5x+4x-20)-(3-(x+7))=0


We calculate terms in parentheses: -(3-(x+7)), so:

3-(x+7)

determiningTheFunctionDomain
-(x+7)+3

We get rid of parentheses

-x-7+3

We add all the numbers together, and all the variables

-1x-4

Back to the equation:

-(-1x-4)


We get rid of parentheses

x^2-5x+4x+1x-20+4=0

We add all the numbers together, and all the variables

x^2-16=0

a = 1; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·1·(-16)
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{-(0)-8}{2*1}=\frac{-8}{2}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*1}=\frac{8}{2}
=4
$