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

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}
$