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4x^2=216
We move all terms to the left:
4x^2-(216)=0
a = 4; b = 0; c = -216;
Δ = b2-4ac
Δ = 02-4·4·(-216)
Δ = 3456
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3456}=\sqrt{576*6}=\sqrt{576}*\sqrt{6}=24\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24\sqrt{6}}{2*4}=\frac{0-24\sqrt{6}}{8} =-\frac{24\sqrt{6}}{8} =-3\sqrt{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24\sqrt{6}}{2*4}=\frac{0+24\sqrt{6}}{8} =\frac{24\sqrt{6}}{8} =3\sqrt{6} $
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