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(2x-12)(4x+43)=9x-26
We move all terms to the left:
(2x-12)(4x+43)-(9x-26)=0
We get rid of parentheses
(2x-12)(4x+43)-9x+26=0
We multiply parentheses ..
(+8x^2+86x-48x-516)-9x+26=0
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} $
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