Worry-able everywhere - 3

Our four cases and their discriminants (from previous problem(s)):

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 second and third equations have positive but non perfect square roots, so they will yield irrational solutions.

Using the quadratic formula on the second equation, $\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2 }$ , gives us the solutions $x=\frac { -1\pm \sqrt { 229 } }{ 2 }$ .

These solutions will make the base $0$ , but we must check to see what they make the powers, ${ 3{ x }^{ 2 }+3\quad and\quad 10x }$ , because if the power(s) is negative, then it would equal $\frac { 1 }{ { 0 }^{ n } }$ , which is undefined.

Substituting in $x=\frac { -1+\sqrt { 229 } }{ 2 }$ ,

$3{ (\frac { -1+\sqrt { 229 } }{ 2 } ) }^{ 2 }+3\cong 24.199$

$10{ (\frac { -1+\sqrt { 229 } }{ 2 } ) }\cong 70.664$

Therefore, this solution works.

Substituting in $x=\frac { -1-\sqrt { 229 } }{ 2 }$ ,

$3{ (\frac { -1-\sqrt { 229 } }{ 2 } ) }^{ 2 }+3\cong 198.199$

$10{ (\frac { -1+\sqrt { 229 } }{ 2 } ) }\cong -80.664$

The $10x$ power becomes negative, and therefore, this is not a valid solution.

Using the quadratic equation on the third equation, we get $x=\frac { -1\pm \sqrt { 233 } }{ 2 }$

Since 1 raised to any power, positive or negative, is 1, these solutions work.

Adding up the solutions, we get $\frac { -1+\sqrt { 233 } }{ 2 } { +\frac { -1-\sqrt { 233 } }{ 2 } +\frac { -1+\sqrt { 229 } }{ 2 } \cong }$ $\boxed {6.066}$