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x^{2}=128
Add 128 to both sides. Anything plus zero gives itself.
x=8\sqrt{2} x=-8\sqrt{2}
Take the square root of both sides of the equation.
x^{2}-128=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\left(-128\right)}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -128 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{0±\sqrt{-4\left(-128\right)}}{2}
Square 0.
x=\frac{0±\sqrt{512}}{2}
Multiply -4 times -128.
x=\frac{0±16\sqrt{2}}{2}
Take the square root of 512.
x=8\sqrt{2}
Now solve the equation x=\frac{0±16\sqrt{2}}{2} when ± is plus.
x=-8\sqrt{2}
Now solve the equation x=\frac{0±16\sqrt{2}}{2} when ± is minus.
x=8\sqrt{2} x=-8\sqrt{2}
The equation is now solved.