x-(x-3)(3+x)=10-(x-2)

Solution for x-(x-3)(3+x)=10-(x-2) equation:

x-(x-3)(3+x)=10-(x-2)

We move all terms to the left:

x-(x-3)(3+x)-(10-(x-2))=0

We add all the numbers together, and all the variables

x-(x-3)(x+3)-(10-(x-2))=0

We use the square of the difference formula

x^2+x-(10-(x-2))+9=0


We calculate terms in parentheses: -(10-(x-2)), so:

10-(x-2)

determiningTheFunctionDomain
-(x-2)+10

We get rid of parentheses

-x+2+10

We add all the numbers together, and all the variables

-1x+12

Back to the equation:

-(-1x+12)


We get rid of parentheses

x^2+x+1x-12+9=0

We add all the numbers together, and all the variables

x^2+2x-3=0

a = 1; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·1·(-3)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*1}=\frac{2}{2}
=1
$