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3x^2-18x+23=3
We move all terms to the left:
3x^2-18x+23-(3)=0
We add all the numbers together, and all the variables
3x^2-18x+20=0
a = 3; b = -18; c = +20;
Δ = b2-4ac
Δ = -182-4·3·20
Δ = 84
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{21}}{2*3}=\frac{18-2\sqrt{21}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{21}}{2*3}=\frac{18+2\sqrt{21}}{6} $
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