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(3x)2-18=24
We move all terms to the left:
(3x)2-18-(24)=0
We add all the numbers together, and all the variables
3x^2-42=0
a = 3; b = 0; c = -42;
Δ = b2-4ac
Δ = 02-4·3·(-42)
Δ = 504
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{14}}{2*3}=\frac{0-6\sqrt{14}}{6} =-\frac{6\sqrt{14}}{6} =-\sqrt{14} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{14}}{2*3}=\frac{0+6\sqrt{14}}{6} =\frac{6\sqrt{14}}{6} =\sqrt{14} $
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