20-12y=(2y-4)(7y+3)

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Solution for 20-12y=(2y-4)(7y+3) equation:



20-12y=(2y-4)(7y+3)
We move all terms to the left:
20-12y-((2y-4)(7y+3))=0
We multiply parentheses ..
-((+14y^2+6y-28y-12))-12y+20=0
We calculate terms in parentheses: -((+14y^2+6y-28y-12)), so:
(+14y^2+6y-28y-12)
We get rid of parentheses
14y^2+6y-28y-12
We add all the numbers together, and all the variables
14y^2-22y-12
Back to the equation:
-(14y^2-22y-12)
We add all the numbers together, and all the variables
-12y-(14y^2-22y-12)+20=0
We get rid of parentheses
-14y^2-12y+22y+12+20=0
We add all the numbers together, and all the variables
-14y^2+10y+32=0
a = -14; b = 10; c = +32;
Δ = b2-4ac
Δ = 102-4·(-14)·32
Δ = 1892
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:
$\sqrt{\Delta}=\sqrt{1892}=\sqrt{4*473}=\sqrt{4}*\sqrt{473}=2\sqrt{473}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{473}}{2*-14}=\frac{-10-2\sqrt{473}}{-28} $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{473}}{2*-14}=\frac{-10+2\sqrt{473}}{-28} $

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