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3(x2)+1=(3x+1)2
We move all terms to the left:
3(x2)+1-((3x+1)2)=0
We add all the numbers together, and all the variables
3x^2-((3x+1)2)+1=0
We calculate terms in parentheses: -((3x+1)2), so:We get rid of parentheses
(3x+1)2
We multiply parentheses
6x+2
Back to the equation:
-(6x+2)
3x^2-6x-2+1=0
We add all the numbers together, and all the variables
3x^2-6x-1=0
a = 3; b = -6; c = -1;
Δ = b2-4ac
Δ = -62-4·3·(-1)
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{3}}{2*3}=\frac{6-4\sqrt{3}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{3}}{2*3}=\frac{6+4\sqrt{3}}{6} $
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