Question 695470
{{{ f(x) = x^2 + 7 }}}
Can you read this OK?
It says: " There is a function called {{{ f(x) }}} which 
equals {{{ x^2 + 7 }}}.
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You can plug in a value for {{{ x }}} on both sides like this:
{{{ f(2) = 2^2 + 7 }}}
{{{ f(2) = 4 + 7 }}}
{{{ f(2) = 11 }}}
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When you say " Solve the problem " , I assume you want
to find the "roots" of the equation. The roots are any and
all values of {{{ x }}} for which {{{ f(x) = 0 }}}
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{{{ f(x) = 0 }}}
{{{ x^2 + 7 = 0 }}}
{{{ x^2 = -7 }}}
Do you understand imaginary numbers? You need to use the fact 
that {{{ i = sqrt(-1) }}}.
{{{ x^2 = -7 }}}
{{{ x^2 = -1*7 }}}
Take the square root of both sides
{{{ x = sqrt(-1) * sqrt(7) }}}
There are two square roots of {{{7}}}. They are 
{{{ sqrt(7) }}} and {{{ -sqrt(7) }}}, another answer is
{{{ x = sqrt(-1) * ( -sqrt(7) ) }}}
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And, since {{{ i = sqrt(-1) }}}, 
{{{ x = sqrt(7)*i }}}
{{{ x = -sqrt(7)*i }}}
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So, you have two imaginary solutions to
{{{ x^2 + 7 = 0 }}}
This is always the case when the plot of the 
parabola ( which the function is ) does not
cross the x-axis.
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If the function {{{ f(x) }}} did cross the x-axis, then
the solutions would be:
{{{ x }}} = value of {{{x}}} at one of the crossings
{{{ x }}} = value of {{{x}}} at the other crossing
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To demonstrate, here's two plots, {{{ f(x) = x^2 + 7 }}}
and a parabola which actually crosses the x-axis.
Your function floats above the x-axis, so it can't
have "real" roots. They must be "imaginary"
{{{ graph( 400, 400, -8, 8, -20, 20, x^2 + 7, x^2 - 7 ) }}} 
Just keep learning how to talk about equations
before solving them. Them key is knowing exactly
what the words mean. 
Good luck