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3x(2x+11)=9
We move all terms to the left:
3x(2x+11)-(9)=0
We multiply parentheses
6x^2+33x-9=0
a = 6; b = 33; c = -9;
Δ = b2-4ac
Δ = 332-4·6·(-9)
Δ = 1305
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1305}=\sqrt{9*145}=\sqrt{9}*\sqrt{145}=3\sqrt{145}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-3\sqrt{145}}{2*6}=\frac{-33-3\sqrt{145}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+3\sqrt{145}}{2*6}=\frac{-33+3\sqrt{145}}{12} $
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