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

Solution for 0=-(x+2)(2x-9) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-(-(+2x^2-9x+4x-18))=0


We calculate terms in parentheses: -(-(+2x^2-9x+4x-18)), so:

-(+2x^2-9x+4x-18)

We get rid of parentheses

-2x^2+9x-4x+18

We add all the numbers together, and all the variables

-2x^2+5x+18

Back to the equation:

-(-2x^2+5x+18)


We get rid of parentheses

2x^2-5x-18=0

a = 2; b = -5; c = -18;
Δ = b2-4ac
Δ = -52-4·2·(-18)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-13}{2*2}=\frac{-8}{4}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+13}{2*2}=\frac{18}{4}
=4+1/2
$