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(6x-11)(3x+2)=5x+1
We move all terms to the left:
(6x-11)(3x+2)-(5x+1)=0
We get rid of parentheses
(6x-11)(3x+2)-5x-1=0
We multiply parentheses ..
(+18x^2+12x-33x-22)-5x-1=0
We get rid of parentheses
18x^2+12x-33x-5x-22-1=0
We add all the numbers together, and all the variables
18x^2-26x-23=0
a = 18; b = -26; c = -23;
Δ = b2-4ac
Δ = -262-4·18·(-23)
Δ = 2332
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2332}=\sqrt{4*583}=\sqrt{4}*\sqrt{583}=2\sqrt{583}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{583}}{2*18}=\frac{26-2\sqrt{583}}{36} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{583}}{2*18}=\frac{26+2\sqrt{583}}{36} $
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