9=(x-3)(2x+1)

Solution for 9=(x-3)(2x+1) equation:

9=(x-3)(2x+1)

We move all terms to the left:

9-((x-3)(2x+1))=0

We multiply parentheses ..

-((+2x^2+x-6x-3))+9=0


We calculate terms in parentheses: -((+2x^2+x-6x-3)), so:

(+2x^2+x-6x-3)

We get rid of parentheses

2x^2+x-6x-3

We add all the numbers together, and all the variables

2x^2-5x-3

Back to the equation:

-(2x^2-5x-3)


We get rid of parentheses

-2x^2+5x+3+9=0

We add all the numbers together, and all the variables

-2x^2+5x+12=0

a = -2; b = 5; c = +12;
Δ = b2-4ac
Δ = 52-4·(-2)·12
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*-2}=\frac{-16}{-4}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*-2}=\frac{6}{-4}
=-1+1/2
$