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(3x+3)(2x-2)=94
We move all terms to the left:
(3x+3)(2x-2)-(94)=0
We multiply parentheses ..
(+6x^2-6x+6x-6)-94=0
We get rid of parentheses
6x^2-6x+6x-6-94=0
We add all the numbers together, and all the variables
6x^2-100=0
a = 6; b = 0; c = -100;
Δ = b2-4ac
Δ = 02-4·6·(-100)
Δ = 2400
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{6}}{2*6}=\frac{0-20\sqrt{6}}{12} =-\frac{20\sqrt{6}}{12} =-\frac{5\sqrt{6}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{6}}{2*6}=\frac{0+20\sqrt{6}}{12} =\frac{20\sqrt{6}}{12} =\frac{5\sqrt{6}}{3} $
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