(3y+2)(11y-32)-8y+34=0

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Solution for (3y+2)(11y-32)-8y+34=0 equation:



(3y+2)(11y-32)-8y+34=0
We add all the numbers together, and all the variables
-8y+(3y+2)(11y-32)+34=0
We multiply parentheses ..
(+33y^2-96y+22y-64)-8y+34=0
We get rid of parentheses
33y^2-96y+22y-8y-64+34=0
We add all the numbers together, and all the variables
33y^2-82y-30=0
a = 33; b = -82; c = -30;
Δ = b2-4ac
Δ = -822-4·33·(-30)
Δ = 10684
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:
$\sqrt{\Delta}=\sqrt{10684}=\sqrt{4*2671}=\sqrt{4}*\sqrt{2671}=2\sqrt{2671}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-82)-2\sqrt{2671}}{2*33}=\frac{82-2\sqrt{2671}}{66} $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-82)+2\sqrt{2671}}{2*33}=\frac{82+2\sqrt{2671}}{66} $

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