3x+3x(2x+5)=33

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Solution for 3x+3x(2x+5)=33 equation:



3x+3x(2x+5)=33
We move all terms to the left:
3x+3x(2x+5)-(33)=0
We multiply parentheses
6x^2+3x+15x-33=0
We add all the numbers together, and all the variables
6x^2+18x-33=0
a = 6; b = 18; c = -33;
Δ = b2-4ac
Δ = 182-4·6·(-33)
Δ = 1116
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:
$\sqrt{\Delta}=\sqrt{1116}=\sqrt{36*31}=\sqrt{36}*\sqrt{31}=6\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{31}}{2*6}=\frac{-18-6\sqrt{31}}{12} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{31}}{2*6}=\frac{-18+6\sqrt{31}}{12} $

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