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3x^2=9x-5
We move all terms to the left:
3x^2-(9x-5)=0
We get rid of parentheses
3x^2-9x+5=0
a = 3; b = -9; c = +5;
Δ = b2-4ac
Δ = -92-4·3·5
Δ = 21
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{21}}{2*3}=\frac{9-\sqrt{21}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{21}}{2*3}=\frac{9+\sqrt{21}}{6} $
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