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(3x)^2=180
We move all terms to the left:
(3x)^2-(180)=0
a = 3; b = 0; c = -180;
Δ = b2-4ac
Δ = 02-4·3·(-180)
Δ = 2160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2160}=\sqrt{144*15}=\sqrt{144}*\sqrt{15}=12\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{15}}{2*3}=\frac{0-12\sqrt{15}}{6} =-\frac{12\sqrt{15}}{6} =-2\sqrt{15} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{15}}{2*3}=\frac{0+12\sqrt{15}}{6} =\frac{12\sqrt{15}}{6} =2\sqrt{15} $
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