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

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Solution for (9+y)(5y-4)=0 equation:



(9+y)(5y-4)=0
We add all the numbers together, and all the variables
(y+9)(5y-4)=0
We multiply parentheses ..
(+5y^2-4y+45y-36)=0
We get rid of parentheses
5y^2-4y+45y-36=0
We add all the numbers together, and all the variables
5y^2+41y-36=0
a = 5; b = 41; c = -36;
Δ = b2-4ac
Δ = 412-4·5·(-36)
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-49}{2*5}=\frac{-90}{10} =-9 $
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+49}{2*5}=\frac{8}{10} =4/5 $

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