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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We use the square of the difference formula

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


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

10-(x-2)2

determiningTheFunctionDomain
-(x-2)2+10

We multiply parentheses

-2x+4+10

We add all the numbers together, and all the variables

-2x+14

Back to the equation:

-(-2x+14)


We get rid of parentheses

x^2+x+2x-14+9=0

We add all the numbers together, and all the variables

x^2+3x-5=0

a = 1; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·1·(-5)
Δ = 29
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{-(3)-\sqrt{29}}{2*1}=\frac{-3-\sqrt{29}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{29}}{2*1}=\frac{-3+\sqrt{29}}{2}
$