How do you solve \[{x^2} - 12x + 52 = 0\]?

Hint: We have a quadratic equation and we can solve this using factorization methods or by using quadratic formulas. In factorization if it’s difficult to split the middle terms we use the quadratic formula or Sridhar’s formula. That is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step by step solution:
Given, \[{x^2} - 12x + 52 = 0\].
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\]. We have \[a = 1\], \[b = - 12\] and \[c = 52\].
For factorization, the standard equation is rewritten as \[a{x^2} + {b_1}x + {b_2}x + c = 0\] such that \[{b_1} \times {b_2} = ac\] and \[{b_1} + {b_2} = b\].
But here we can’t solve by factorization,
So we use quadratic formula to solve this,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substituting we have,
\[ \Rightarrow x = \dfrac{{ - \left( { - 12} \right) \pm \sqrt {{{\left( { - 12} \right)}^2} - 4\left( 1 \right)\left( {52} \right)} }}{{2\left( 1 \right)}}\]
\[ \Rightarrow x = \dfrac{{12 \pm \sqrt {144 - 208} }}{2}\]
\[ \Rightarrow x = \dfrac{{12 \pm \sqrt { - 64} }}{2}\]
But we know that \[\sqrt { - 1} = i\] and 64 is a perfect square the,
\[ \Rightarrow x = \dfrac{{12 \pm i\sqrt {64} }}{2}\]
\[ \Rightarrow x = \dfrac{{12 \pm 8i}}{2}\]
Taking 2 common we have,
\[ \Rightarrow x = \dfrac{{2\left( {6 \pm 4i} \right)}}{2}\]
\[ \Rightarrow x = 6 \pm 4i\]

Thus we have two roots \[ x = 6 + 4i\] and \[ x = 6 - 4i\].

Note: The highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis, that is the roots are simply the x-intercepts.