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10x^2=540
We move all terms to the left:
10x^2-(540)=0
a = 10; b = 0; c = -540;
Δ = b2-4ac
Δ = 02-4·10·(-540)
Δ = 21600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{21600}=\sqrt{3600*6}=\sqrt{3600}*\sqrt{6}=60\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-60\sqrt{6}}{2*10}=\frac{0-60\sqrt{6}}{20} =-\frac{60\sqrt{6}}{20} =-3\sqrt{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+60\sqrt{6}}{2*10}=\frac{0+60\sqrt{6}}{20} =\frac{60\sqrt{6}}{20} =3\sqrt{6} $
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