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(9x+4)/x^2+5=0
Domain of the equation: x^2!=0We multiply all the terms by the denominator
x^2!=0/
x^2!=√0
x!=0
x∈R
(9x+4)+5*x^2=0
We add all the numbers together, and all the variables
5x^2+(9x+4)=0
We get rid of parentheses
5x^2+9x+4=0
a = 5; b = 9; c = +4;
Δ = b2-4ac
Δ = 92-4·5·4
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-1}{2*5}=\frac{-10}{10} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+1}{2*5}=\frac{-8}{10} =-4/5 $
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