(2x-12)(9x-26)=(4x+43)

Solution for (2x-12)(9x-26)=(4x+43) equation:

(2x-12)(9x-26)=(4x+43)

We move all terms to the left:

(2x-12)(9x-26)-((4x+43))=0

We multiply parentheses ..

(+18x^2-52x-108x+312)-((4x+43))=0


We calculate terms in parentheses: -((4x+43)), so:

(4x+43)

We get rid of parentheses

4x+43

Back to the equation:

-(4x+43)


We get rid of parentheses

18x^2-52x-108x-4x+312-43=0

We add all the numbers together, and all the variables

18x^2-164x+269=0

a = 18; b = -164; c = +269;
Δ = b2-4ac
Δ = -1642-4·18·269
Δ = 7528
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:

$\sqrt{\Delta}=\sqrt{7528}=\sqrt{4*1882}=\sqrt{4}*\sqrt{1882}=2\sqrt{1882}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-164)-2\sqrt{1882}}{2*18}=\frac{164-2\sqrt{1882}}{36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-164)+2\sqrt{1882}}{2*18}=\frac{164+2\sqrt{1882}}{36}
$