0=(x-5)(2x-9)

Solution for 0=(x-5)(2x-9) equation:

0=(x-5)(2x-9)

We move all terms to the left:

0-((x-5)(2x-9))=0

We add all the numbers together, and all the variables

-((x-5)(2x-9))=0

We multiply parentheses ..

-((+2x^2-9x-10x+45))=0


We calculate terms in parentheses: -((+2x^2-9x-10x+45)), so:

(+2x^2-9x-10x+45)

We get rid of parentheses

2x^2-9x-10x+45

We add all the numbers together, and all the variables

2x^2-19x+45

Back to the equation:

-(2x^2-19x+45)


We get rid of parentheses

-2x^2+19x-45=0

a = -2; b = 19; c = -45;
Δ = b2-4ac
Δ = 192-4·(-2)·(-45)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-1}{2*-2}=\frac{-20}{-4}
=+5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+1}{2*-2}=\frac{-18}{-4}
=4+1/2
$