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(y+3)(y+5)=0
We multiply parentheses ..
(+y^2+5y+3y+15)=0
We get rid of parentheses
y^2+5y+3y+15=0
We add all the numbers together, and all the variables
y^2+8y+15=0
a = 1; b = 8; c = +15;
Δ = b2-4ac
Δ = 82-4·1·15
Δ = 4
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{4}=2$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2}{2*1}=\frac{-10}{2} =-5 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*1}=\frac{-6}{2} =-3 $
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