2x-5=(x^2)+(13-9x)

Solution for 2x-5=(x^2)+(13-9x) equation:

2x-5=(x^2)+(13-9x)

We move all terms to the left:

2x-5-((x^2)+(13-9x))=0

We add all the numbers together, and all the variables

2x-(x^2+(-9x+13))-5=0


We calculate terms in parentheses: -(x^2+(-9x+13)), so:

x^2+(-9x+13)

We get rid of parentheses

x^2-9x+13

Back to the equation:

-(x^2-9x+13)


We get rid of parentheses

-x^2+2x+9x-13-5=0

We add all the numbers together, and all the variables

-1x^2+11x-18=0

a = -1; b = 11; c = -18;
Δ = b2-4ac
Δ = 112-4·(-1)·(-18)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-7}{2*-1}=\frac{-18}{-2}
=+9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+7}{2*-1}=\frac{-4}{-2}
=+2
$