(3y+1)(8-y)=0

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Solution for (3y+1)(8-y)=0 equation:



(3y+1)(8-y)=0
We add all the numbers together, and all the variables
(3y+1)(-1y+8)=0
We multiply parentheses ..
(-3y^2+24y-1y+8)=0
We get rid of parentheses
-3y^2+24y-1y+8=0
We add all the numbers together, and all the variables
-3y^2+23y+8=0
a = -3; b = 23; c = +8;
Δ = b2-4ac
Δ = 232-4·(-3)·8
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-25}{2*-3}=\frac{-48}{-6} =+8 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+25}{2*-3}=\frac{2}{-6} =-1/3 $

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