32=y(y+4)

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Solution for 32=y(y+4) equation:



32=y(y+4)
We move all terms to the left:
32-(y(y+4))=0
We calculate terms in parentheses: -(y(y+4)), so:
y(y+4)
We multiply parentheses
y^2+4y
Back to the equation:
-(y^2+4y)
We get rid of parentheses
-y^2-4y+32=0
We add all the numbers together, and all the variables
-1y^2-4y+32=0
a = -1; b = -4; c = +32;
Δ = b2-4ac
Δ = -42-4·(-1)·32
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*-1}=\frac{-8}{-2} =+4 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*-1}=\frac{16}{-2} =-8 $

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