Z tables normal distribution $P(10 - c < X < 10 + c) = 0.95$. Find c. For N(10, 2)

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$P(10 - c < X < 10 + c) = 0.95$. Find c

attempt:

$P(10 - c < X < 10 + c) = \phi(\frac{10 + c - 10}{\sqrt{2}}) - \phi(\frac{10-c-10}{\sqrt{2}}) = \phi(\frac{c}{\sqrt{2}}) - (1 - \phi(\frac{c}{\sqrt{2}})) = 2\phi(\frac{c}{\sqrt{2}}) - 1$

$0.975 = \phi(\frac{c}{\sqrt{2}})$

not sure how to simplify to $\phi$. I know $\phi(0.975) \approx 0.834$ according to Z table. How do I compute the inverse?

$\phi^{-1}(0.975) = \frac{c}{\sqrt{2}}$

using the Z table.