9x-(1x-36)=3x(-7+x)

Solution for 9x-(1x-36)=3x(-7+x) equation:

9x-(1x-36)=3x(-7+x)

We move all terms to the left:

9x-(1x-36)-(3x(-7+x))=0

We add all the numbers together, and all the variables

9x-(x-36)-(3x(x-7))=0

We get rid of parentheses

9x-x-(3x(x-7))+36=0


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

3x(x-7)

We multiply parentheses

3x^2-21x

Back to the equation:

-(3x^2-21x)


We add all the numbers together, and all the variables

8x-(3x^2-21x)+36=0

We get rid of parentheses

-3x^2+8x+21x+36=0

We add all the numbers together, and all the variables

-3x^2+29x+36=0

a = -3; b = 29; c = +36;
Δ = b2-4ac
Δ = 292-4·(-3)·36
Δ = 1273
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{-(29)-\sqrt{1273}}{2*-3}=\frac{-29-\sqrt{1273}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{1273}}{2*-3}=\frac{-29+\sqrt{1273}}{-6}
$