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(3x+3)2x=146
We move all terms to the left:
(3x+3)2x-(146)=0
We multiply parentheses
6x^2+6x-146=0
a = 6; b = 6; c = -146;
Δ = b2-4ac
Δ = 62-4·6·(-146)
Δ = 3540
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3540}=\sqrt{4*885}=\sqrt{4}*\sqrt{885}=2\sqrt{885}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{885}}{2*6}=\frac{-6-2\sqrt{885}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{885}}{2*6}=\frac{-6+2\sqrt{885}}{12} $
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