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

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Solution for (5+y)(3y-8)=0 equation:



(5+y)(3y-8)=0
We add all the numbers together, and all the variables
(y+5)(3y-8)=0
We multiply parentheses ..
(+3y^2-8y+15y-40)=0
We get rid of parentheses
3y^2-8y+15y-40=0
We add all the numbers together, and all the variables
3y^2+7y-40=0
a = 3; b = 7; c = -40;
Δ = b2-4ac
Δ = 72-4·3·(-40)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-23}{2*3}=\frac{-30}{6} =-5 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+23}{2*3}=\frac{16}{6} =2+2/3 $

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