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3x(3x-4)+1=0
We multiply parentheses
9x^2-12x+1=0
a = 9; b = -12; c = +1;
Δ = b2-4ac
Δ = -122-4·9·1
Δ = 108
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{3}}{2*9}=\frac{12-6\sqrt{3}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{3}}{2*9}=\frac{12+6\sqrt{3}}{18} $
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