Respuesta :
[tex] \bf \stackrel{\textit{solving for \underline{x}}}{12x^2-48=0\implies 12x^2=48\implies x^2=\cfrac{48}{12}\implies x^2=4}
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x=\pm\sqrt{4}\implies x=\pm 2 [/tex]
All three methods are easy and effective. Here's solving the equation using all three methods.
First Option:
[tex] 12x^2-48=0\\ 12x^2=48\\ x^2=4\\ x=+/-\sqrt{4} \\ x=2,-2 [/tex]
Third Option (Remember that the equation for the quadratic formula is [tex] x=\frac{-b+\sqrt{b^2-4ac}}{2a},\frac{-b-\sqrt{b^2-4ac}}{2a} [/tex] , with a = x^2 coefficient, b = x coefficient, and c = constant):
[tex] x=\frac{-0+\sqrt{0^2-4*12*(-48)}}{2*12},\frac{-0-\sqrt{0^2-4*12*(-48)}}{2*12}\\\\ x=\frac{-0+\sqrt{2304}}{24},\frac{-0-\sqrt{2304}}{24}\\ \\ x=\frac{48}{24},\frac{-48}{24}\\\\ x=2,-2 [/tex]
Fourth Option:
[tex] 12x^2-48=0\\12(x^2-4)=0 \\ 12(x+2)(x-2)=0\\ \\ x+2=0\\ x=-2\\ \\ x-2=0\\ x=2 [/tex]