9x^2=9-x(43+x)

Solution for 9x^2=9-x(43+x) equation:

9x^2=9-x(43+x)

We move all terms to the left:

9x^2-(9-x(43+x))=0

We add all the numbers together, and all the variables

9x^2-(9-x(x+43))=0


We calculate terms in parentheses: -(9-x(x+43)), so:

9-x(x+43)

determiningTheFunctionDomain
-x(x+43)+9

We multiply parentheses

-x^2-43x+9

We add all the numbers together, and all the variables

-1x^2-43x+9

Back to the equation:

-(-1x^2-43x+9)


We get rid of parentheses

9x^2+1x^2+43x-9=0

We add all the numbers together, and all the variables

10x^2+43x-9=0

a = 10; b = 43; c = -9;
Δ = b2-4ac
Δ = 432-4·10·(-9)
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-47}{2*10}=\frac{-90}{20}
=-4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+47}{2*10}=\frac{4}{20}
=1/5
$