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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+8x^2+86x-48x-516)-((9x-26))=0


We calculate terms in parentheses: -((9x-26)), so:

(9x-26)

We get rid of parentheses

9x-26

Back to the equation:

-(9x-26)


We get rid of parentheses

8x^2+86x-48x-9x-516+26=0

We add all the numbers together, and all the variables

8x^2+29x-490=0

a = 8; b = 29; c = -490;
Δ = b2-4ac
Δ = 292-4·8·(-490)
Δ = 16521
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-\sqrt{16521}}{2*8}=\frac{-29-\sqrt{16521}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+\sqrt{16521}}{2*8}=\frac{-29+\sqrt{16521}}{16}
$