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3x^2-36x+324=282
We move all terms to the left:
3x^2-36x+324-(282)=0
We add all the numbers together, and all the variables
3x^2-36x+42=0
a = 3; b = -36; c = +42;
Δ = b2-4ac
Δ = -362-4·3·42
Δ = 792
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{792}=\sqrt{36*22}=\sqrt{36}*\sqrt{22}=6\sqrt{22}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-6\sqrt{22}}{2*3}=\frac{36-6\sqrt{22}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+6\sqrt{22}}{2*3}=\frac{36+6\sqrt{22}}{6} $
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