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3x^2+5=9x+6
We move all terms to the left:
3x^2+5-(9x+6)=0
We get rid of parentheses
3x^2-9x-6+5=0
We add all the numbers together, and all the variables
3x^2-9x-1=0
a = 3; b = -9; c = -1;
Δ = b2-4ac
Δ = -92-4·3·(-1)
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{-(-9)-\sqrt{93}}{2*3}=\frac{9-\sqrt{93}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{93}}{2*3}=\frac{9+\sqrt{93}}{6} $
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