(4+y)(5y-2)=0

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Solution for (4+y)(5y-2)=0 equation:



(4+y)(5y-2)=0
We add all the numbers together, and all the variables
(y+4)(5y-2)=0
We multiply parentheses ..
(+5y^2-2y+20y-8)=0
We get rid of parentheses
5y^2-2y+20y-8=0
We add all the numbers together, and all the variables
5y^2+18y-8=0
a = 5; b = 18; c = -8;
Δ = b2-4ac
Δ = 182-4·5·(-8)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*5}=\frac{-40}{10} =-4 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*5}=\frac{4}{10} =2/5 $

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