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3x^2-24x=-39
We move all terms to the left:
3x^2-24x-(-39)=0
We add all the numbers together, and all the variables
3x^2-24x+39=0
a = 3; b = -24; c = +39;
Δ = b2-4ac
Δ = -242-4·3·39
Δ = 108
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-6\sqrt{3}}{2*3}=\frac{24-6\sqrt{3}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+6\sqrt{3}}{2*3}=\frac{24+6\sqrt{3}}{6} $
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