Solve for x
x=\sqrt{3}\approx 1.732050808
x=-\sqrt{3}\approx -1.732050808
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4x^{2}=12
Add 12 to both sides. Anything plus zero gives itself.
x^{2}=\frac{12}{4}
Divide both sides by 4.
x^{2}=3
Divide 12 by 4 to get 3.
x=\sqrt{3} x=-\sqrt{3}
Take the square root of both sides of the equation.
4x^{2}-12=0
Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.
x=\frac{0±\sqrt{0^{2}-4\times 4\left(-12\right)}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 0 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{0±\sqrt{-4\times 4\left(-12\right)}}{2\times 4}
Square 0.
x=\frac{0±\sqrt{-16\left(-12\right)}}{2\times 4}
Multiply -4 times 4.
x=\frac{0±\sqrt{192}}{2\times 4}
Multiply -16 times -12.
x=\frac{0±8\sqrt{3}}{2\times 4}
Take the square root of 192.
x=\frac{0±8\sqrt{3}}{8}
Multiply 2 times 4.
x=\sqrt{3}
Now solve the equation x=\frac{0±8\sqrt{3}}{8} when ± is plus.
x=-\sqrt{3}
Now solve the equation x=\frac{0±8\sqrt{3}}{8} when ± is minus.
x=\sqrt{3} x=-\sqrt{3}
The equation is now solved.
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