(4y+3)(8+y)=0

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Solution for (4y+3)(8+y)=0 equation:



(4y+3)(8+y)=0
We add all the numbers together, and all the variables
(4y+3)(y+8)=0
We multiply parentheses ..
(+4y^2+32y+3y+24)=0
We get rid of parentheses
4y^2+32y+3y+24=0
We add all the numbers together, and all the variables
4y^2+35y+24=0
a = 4; b = 35; c = +24;
Δ = b2-4ac
Δ = 352-4·4·24
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-29}{2*4}=\frac{-64}{8} =-8 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+29}{2*4}=\frac{-6}{8} =-3/4 $

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