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3x(3x+4)=2
We move all terms to the left:
3x(3x+4)-(2)=0
We multiply parentheses
9x^2+12x-2=0
a = 9; b = 12; c = -2;
Δ = b2-4ac
Δ = 122-4·9·(-2)
Δ = 216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{6}}{2*9}=\frac{-12-6\sqrt{6}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{6}}{2*9}=\frac{-12+6\sqrt{6}}{18} $
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