Question 1075729
The quadratic formula says that the roots are:
{{{ x = ( -b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
when the equation has the form:
{{{ y = a*x^2 + b*x + c }}}
They decided that {{{ b = a }}} ( a little confusing )
so the list is:
{{{ a = 4 }}}
{{{ b = a }}}
{{{ c = 25 }}}
Now I can say:
{{{ x = ( -a +- sqrt( a^2-4*4*25 ))/(2*4) }}} 
{{{ x = ( -a +- sqrt( a^2 - 400 ))/8 }}} 
The roots will be non-real ( imaginary ) whenever
{{{ a^2 - 400 }}} is negative. ( this is called the discriminant )
So I can say:
{{{ a^2 - 400 < 0 }}}
{{{ a^2 < 400 }}}
{{{ a < 20 }}}
and also:
{{{ a > -20 }}}
Putting these together:
{{{ -20 < a < 20  }}}
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example:
{{{ a = 0 }}} satisfies the conditions, so
{{{ x = ( -a +- sqrt( a^2 - 400 ))/8 }}} 
{{{ x = ( -0 +- sqrt( 0^2 - 400 ))/8 }}} 
{{{ x = (1/8)*( sqrt( -400 ) ) }}}
{{{ x = (1/8)*( 20i ) }}}
{{{ x = (5/2)*i }}}
and also:
{{{ x = (1/8)*( -sqrt( -400 ) ) }}}
{{{ x = (-5/2)*i }}}
Here is the plot when {{{ a = 0 }}}
{{{ graph( 400, 400, -6, 6, -5, 50, 4x^2 + 25 ) }}}