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3x^2+6x=3x+7
We move all terms to the left:
3x^2+6x-(3x+7)=0
We get rid of parentheses
3x^2+6x-3x-7=0
We add all the numbers together, and all the variables
3x^2+3x-7=0
a = 3; b = 3; c = -7;
Δ = b2-4ac
Δ = 32-4·3·(-7)
Δ = 93
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{93}}{2*3}=\frac{-3-\sqrt{93}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{93}}{2*3}=\frac{-3+\sqrt{93}}{6} $
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