How smart are you at equations ?

After simplifying the equation, you get the following quadratic equation -

$3{ x }^{ 2 }-100x+950=0$

Then apply the formula to figure out the number of real roots -

$\sqrt { { b }^{ 2 }-4ac } \\ =\sqrt { { 100 }^{ 2 }-4\cdot 3\cdot 950 } \\ =\sqrt { 10000-11400 } \\ =\sqrt { -1400 }$

which clearly means that the equation has no real roots

Sum of non-negative numbers can be $0$ only if all are simultaneously $0$ , which isn't the case here.

The given equation can be written as $\color{#3D99F6}{(x-10)^2+(x-15)^2+(25-x)^2=0}$ which is of the form ${ a }^{ 2 }+b^{ 2 }+c^{ 2 }$ $\therefore$ ${ x }^{ 2 }=2\left[ (x-10)(x-15)+(x-15)(25-x)+(x-10)(25-x) \right] \\ \quad =2\left[ 50x-475-{ x }^{ 2 } \right] \\ 3{ x }^{ 2 }-100x+950=0$

There exists no real solution for the above equation.

Since it's quadratic equation then the answer is always 0 or positive integer. But there are 3 quadratic equations while only one of them can be zero and the others are positive integer. So the sum of the 3 quadratic equations is impossible 0.

We know x^2>=0 , so each term on LHS is non negative.Now their summation is 0=> each term on LHS=0 => possible values of x =10,15,25 .But cross verification rejects all these values (e.g at x=10 remaining two terms make LHS positive).Hence no real value of 'x' satisfies the given equation.