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3x^2+60=180
We move all terms to the left:
3x^2+60-(180)=0
We add all the numbers together, and all the variables
3x^2-120=0
a = 3; b = 0; c = -120;
Δ = b2-4ac
Δ = 02-4·3·(-120)
Δ = 1440
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1440}=\sqrt{144*10}=\sqrt{144}*\sqrt{10}=12\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{10}}{2*3}=\frac{0-12\sqrt{10}}{6} =-\frac{12\sqrt{10}}{6} =-2\sqrt{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{10}}{2*3}=\frac{0+12\sqrt{10}}{6} =\frac{12\sqrt{10}}{6} =2\sqrt{10} $
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