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2x(3x-3)+7(1+6x)=0
We add all the numbers together, and all the variables
2x(3x-3)+7(6x+1)=0
We multiply parentheses
6x^2-6x+42x+7=0
We add all the numbers together, and all the variables
6x^2+36x+7=0
a = 6; b = 36; c = +7;
Δ = b2-4ac
Δ = 362-4·6·7
Δ = 1128
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1128}=\sqrt{4*282}=\sqrt{4}*\sqrt{282}=2\sqrt{282}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{282}}{2*6}=\frac{-36-2\sqrt{282}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{282}}{2*6}=\frac{-36+2\sqrt{282}}{12} $
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