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5a^2+3a+1=3a^2
We move all terms to the left:
5a^2+3a+1-(3a^2)=0
determiningTheFunctionDomain 5a^2-3a^2+3a+1=0
We add all the numbers together, and all the variables
2a^2+3a+1=0
a = 2; b = 3; c = +1;
Δ = b2-4ac
Δ = 32-4·2·1
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-1}{2*2}=\frac{-4}{4} =-1 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+1}{2*2}=\frac{-2}{4} =-1/2 $
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