9x-57=(x-3)(x-5)

Solution for 9x-57=(x-3)(x-5) equation:

9x-57=(x-3)(x-5)

We move all terms to the left:

9x-57-((x-3)(x-5))=0

We multiply parentheses ..

-((+x^2-5x-3x+15))+9x-57=0


We calculate terms in parentheses: -((+x^2-5x-3x+15)), so:

(+x^2-5x-3x+15)

We get rid of parentheses

x^2-5x-3x+15

We add all the numbers together, and all the variables

x^2-8x+15

Back to the equation:

-(x^2-8x+15)


We add all the numbers together, and all the variables

9x-(x^2-8x+15)-57=0

We get rid of parentheses

-x^2+9x+8x-15-57=0

We add all the numbers together, and all the variables

-1x^2+17x-72=0

a = -1; b = 17; c = -72;
Δ = b2-4ac
Δ = 172-4·(-1)·(-72)
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{-(17)-1}{2*-1}=\frac{-18}{-2}
=+9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+1}{2*-1}=\frac{-16}{-2}
=+8
$