Solve \[{x^2} = 150.\]

Hint: To solve for \[x\], rearrange the given equation and express it as a difference of two squares. Then use the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\], to simplify the equations and then equate the terms to find the value of \[x\].

Complete step by step solution:
To solve for \[x\] in the given expression, first bring all the terms to one side of the equation and try to express it as a difference of two squares.
\[{x^2} = 150.\]
\[ \Rightarrow {x^2} - 150 = 0\]
\[ \Rightarrow {x^2} - {\left( {5\sqrt 6 } \right)^2} = 0\]
Now use the formula \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\],
\[ \Rightarrow \left( {x + 5\sqrt 6 } \right)\left( {x - 5\sqrt 6 } \right) = 0\]
Now if \[\left( {a + b} \right)\left( {a - b} \right) = 0\], then either \[\left( {a + b} \right) = 0\] or \[\left( {a - b} \right) = 0\].
\[ \Rightarrow \left( {x + 5\sqrt 6 } \right) = 0\] or \[ \Rightarrow \left( {x - 5\sqrt 6 } \right) = 0\]
\[ \Rightarrow x = 5\sqrt 6 \] or \[x = - 5\sqrt 6 \].
Now \[\sqrt 6 = \sqrt 3 \times \sqrt 2 \],
\[ \Rightarrow \sqrt 6 = 1.732 \times 1.414\]
\[\sqrt 6 = 2.449\]
\[\therefore x = 5 \times 2.449 = 12.24\] or \[\therefore x = - \left( {5 \times 2.449} \right) = - 12.24\].

additional information:
This equation represents a quadratic equation, hence can be solved by using the Sridharacharya method.
According to that for a quadratic equation, \[a{x^2} + bx + c = 0\], \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Our equation is \[{x^2} - 150 = 0\], \[\therefore a = 1,b = 0,c = - 150\]
\[\therefore x = \dfrac{{ - 0 \pm \sqrt {{0^2} - 4 \cdot 1 \cdot \left( { - 150} \right)} }}{{2 \cdot 1}}\]
\[\therefore x = \dfrac{{ - 0 \pm \sqrt {600} }}{2}\]
\[\therefore x = \dfrac{{ \pm 10\sqrt 6 }}{2}\]
\[\therefore x = \pm 5\sqrt 6 \].

Note: The nature of roots of any quadratic equation, \[a{x^2} + bx + c = 0\] depends on its discriminant(D) where \[D = \sqrt {{b^2} - 4ac} \].
If \[D > 0\], Roots are real and unequal.
If \[D < 0\], Roots are unreal
If \[D = 0\], roots are real and equal.
Also while solving these problems remember that any quadratic equation always has two solutions.