(9x-31)(6x-2)=(3x+33)

Solution for (9x-31)(6x-2)=(3x+33) equation:

(9x-31)(6x-2)=(3x+33)

We move all terms to the left:

(9x-31)(6x-2)-((3x+33))=0

We multiply parentheses ..

(+54x^2-18x-186x+62)-((3x+33))=0


We calculate terms in parentheses: -((3x+33)), so:

(3x+33)

We get rid of parentheses

3x+33

Back to the equation:

-(3x+33)


We get rid of parentheses

54x^2-18x-186x-3x+62-33=0

We add all the numbers together, and all the variables

54x^2-207x+29=0

a = 54; b = -207; c = +29;
Δ = b2-4ac
Δ = -2072-4·54·29
Δ = 36585
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{36585}=\sqrt{9*4065}=\sqrt{9}*\sqrt{4065}=3\sqrt{4065}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-207)-3\sqrt{4065}}{2*54}=\frac{207-3\sqrt{4065}}{108}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-207)+3\sqrt{4065}}{2*54}=\frac{207+3\sqrt{4065}}{108}
$