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2 answers:
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Let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3
Then the equation is (x-x₁)(x-x₂) = 0
If you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.
You can see that a=1 (there is just an x²)
b = (-5 -2√7 + 2√7 - 5)/3 = -10/3
c = ((-5-2√7)/3 ) * (-5+2√7)/3) = -1/3
So the equation you're looking for is:
Then the equation is (x-x₁)(x-x₂) = 0
If you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.
You can see that a=1 (there is just an x²)
b = (-5 -2√7 + 2√7 - 5)/3 = -10/3
c = ((-5-2√7)/3 ) * (-5+2√7)/3) = -1/3
So the equation you're looking for is: