when $\displaystyle c=-7 $and $\displaystyle d=4$
$\displaystyle \sqrt{d} - dc^2$
$\displaystyle \sqrt{4} - 4 x (-7)^2$
$\displaystyle 2 - 4 x (-7)^2 = -194$
this right?
did you know that
$\displaystyle (-2)^2=4$
$\displaystyle (-4)^2=16$
$\displaystyle (-7)^2=49$
so
$\displaystyle \sqrt{4}=\mp 2 $
$\displaystyle \sqrt{16}=\mp 4 $
$\displaystyle \sqrt{49}=\mp 7 $
in general
$\displaystyle \sqrt{a^2}=\mp a $ a is real number
in you question
$\displaystyle \sqrt{4}-4(-7)^2$ have two solutions since $\displaystyle \sqrt{4}=2 $ and $\displaystyle \sqrt{4}=-2$
first one
$\displaystyle 2-4(49)=2-196=-194$
second one
$\displaystyle -2-4(49)=-2-196=-198$
One might say that the evaluation has two answers, but this would be incorrect.
While one solves "x^2 = 4" for the two solution values, -2 and +2, "the" value of the square root of 4, sqrt[4], is defined to be the principal (that is, the positive real) value: sqrt[4] = 2.
So no, there will most definitely not be two answers to this one evaluation exercise!
$\displaystyle \sqrt{d}\, -\, dc^2\, =\,\sqrt{4}\, -\, (4)(-7)^2$
. . .$\displaystyle =\, 2\, -\,4(49)\, =\, 2\, -\, 196\, =\, -194$
And the above is the only answer!