Worry-able everywhere - 2

Looking back on our 4 cases and the discriminant of each from the solution to the first problem:

Case 1: $3{ x }^{ 2 }+3=10x$

Since both sides have the same base, if the powers are equal, the two sides will equal each other.

Discriminant: $64$

Case 2: ${ x }^{ 2 }+x-57=0$

Zero to any positive power will equal zero, so both sides will equal regardless of having different powers.

Discriminant: $229$

Case 3: ${ x }^{ 2 }+x-57=1$

Similarly, 1 to any power will equal 1.

Discriminant: $233$

Case 4: ${ x }^{ 2 }+x-57=-1$

As long as the powers on both sides are either both even or both odd, -1 to a power will equal 1 and -1, respectively.

Discriminant: $225$

We see that the first case and the fourth case yield discriminants that are perfect squares, and will therefore yield rational solutions.

Factoring the first equation gives us

$(3x-1)(x-3)=0$

So, $x=3,x=\frac { 1 }{ 3 }$

Substituting these values in the original problem, we see that they both work.

Factoring the fourth equation gives us

$(x-7)(x+8)=0$

So, $x=7,\quad x=-8$

However, since this case sets the base as $-1$ , we must check if these values make both $3{ x }^{ 2 }+3$ and $10x$ even or both odd.

Substituting $x=7$ , we get $3{ (7) }^{ 2 }+3=150,\quad 10(7)=70$

These are both even, so the end equation is $1=1$

Substituting $x=-8$ , we get $3{ (-8) }^{ 2 }+3=195,\quad 10(-8)=-80$

Since one is odd and one is even, the end result in $-1=1$ , which is not true.

We disregard the solution $x=-8$ , so adding up the other solutions gives us $3+\frac { 1 }{ 3 } +7=10\frac { 1 }{ 3 } ,\quad or\quad 10.\overline { 3 }$

Rounding that off gives our answer, $\boxed {10.333}$