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(2x^2+10x+8)/(x+4)=0
Domain of the equation: (x+4)!=0We multiply all the terms by the denominator
We move all terms containing x to the left, all other terms to the right
x!=-4
x∈R
(2x^2+10x+8)=0
We get rid of parentheses
2x^2+10x+8=0
a = 2; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·2·8
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*2}=\frac{-16}{4} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*2}=\frac{-4}{4} =-1 $
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