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8x^{2}=1
Add 1 to both sides. Anything plus zero gives itself.
x^{2}=\frac{1}{8}
Divide both sides by 8.
x=\frac{\sqrt{2}}{4} x=-\frac{\sqrt{2}}{4}
Take the square root of both sides of the equation.
8x^{2}-1=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 8\left(-1\right)}}{2\times 8}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 0 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{0±\sqrt{-4\times 8\left(-1\right)}}{2\times 8}
Square 0.
x=\frac{0±\sqrt{-32\left(-1\right)}}{2\times 8}
Multiply -4 times 8.
x=\frac{0±\sqrt{32}}{2\times 8}
Multiply -32 times -1.
x=\frac{0±4\sqrt{2}}{2\times 8}
Take the square root of 32.
x=\frac{0±4\sqrt{2}}{16}
Multiply 2 times 8.
x=\frac{\sqrt{2}}{4}
Now solve the equation x=\frac{0±4\sqrt{2}}{16} when ± is plus.
x=-\frac{\sqrt{2}}{4}
Now solve the equation x=\frac{0±4\sqrt{2}}{16} when ± is minus.
x=\frac{\sqrt{2}}{4} x=-\frac{\sqrt{2}}{4}
The equation is now solved.