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(6x^2+16x+8)/(3x+2)=0
Domain of the equation: (3x+2)!=0We multiply all the terms by the denominator
We move all terms containing x to the left, all other terms to the right
3x!=-2
x!=-2/3
x!=-2/3
x∈R
(6x^2+16x+8)=0
We get rid of parentheses
6x^2+16x+8=0
a = 6; b = 16; c = +8;
Δ = b2-4ac
Δ = 162-4·6·8
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8}{2*6}=\frac{-24}{12} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8}{2*6}=\frac{-8}{12} =-2/3 $
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