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(x+7)(x+3)/(x+5)^2=0
Domain of the equation: (x+5)^2!=0We multiply parentheses ..
x∈R
(+x^2+3x+7x+21)/(x+5)^2=0
We multiply all the terms by the denominator
(+x^2+3x+7x+21)=0
We get rid of parentheses
x^2+3x+7x+21=0
We add all the numbers together, and all the variables
x^2+10x+21=0
a = 1; b = 10; c = +21;
Δ = b2-4ac
Δ = 102-4·1·21
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4}{2*1}=\frac{-14}{2} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4}{2*1}=\frac{-6}{2} =-3 $
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