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3x^2-24x+23=0
a = 3; b = -24; c = +23;
Δ = b2-4ac
Δ = -242-4·3·23
Δ = 300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-10\sqrt{3}}{2*3}=\frac{24-10\sqrt{3}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+10\sqrt{3}}{2*3}=\frac{24+10\sqrt{3}}{6} $
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