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(3x+3)3x=36
We move all terms to the left:
(3x+3)3x-(36)=0
We multiply parentheses
9x^2+9x-36=0
a = 9; b = 9; c = -36;
Δ = b2-4ac
Δ = 92-4·9·(-36)
Δ = 1377
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1377}=\sqrt{81*17}=\sqrt{81}*\sqrt{17}=9\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9\sqrt{17}}{2*9}=\frac{-9-9\sqrt{17}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9\sqrt{17}}{2*9}=\frac{-9+9\sqrt{17}}{18} $
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