9x-(1x-36)=1x(-7+x)

Solution for 9x-(1x-36)=1x(-7+x) equation:

9x-(1x-36)=1x(-7+x)

We move all terms to the left:

9x-(1x-36)-(1x(-7+x))=0

We add all the numbers together, and all the variables

9x-(x-36)-(1x(x-7))=0

We get rid of parentheses

9x-x-(1x(x-7))+36=0


We calculate terms in parentheses: -(1x(x-7)), so:

1x(x-7)

We multiply parentheses

x^2-7x

Back to the equation:

-(x^2-7x)


We add all the numbers together, and all the variables

8x-(x^2-7x)+36=0

We get rid of parentheses

-x^2+8x+7x+36=0

We add all the numbers together, and all the variables

-1x^2+15x+36=0

a = -1; b = 15; c = +36;
Δ = b2-4ac
Δ = 152-4·(-1)·36
Δ = 369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{41}}{2*-1}=\frac{-15-3\sqrt{41}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{41}}{2*-1}=\frac{-15+3\sqrt{41}}{-2}
$