Answer:
[tex]x=2$\pm$\sqrt{42}[/tex]
Step-by-step explanation:
The given equation is:
[tex]x^{2} -4x-9=29\\\Rightarrow x^{2} -4x-9-29=0\\\Rightarrow x^{2} -4x-38=0[/tex]
Formula:
A quadratic equation [tex]ax^{2} +bx+c=0[/tex] has the following roots:
[tex]x=\dfrac{-b+\sqrt D}{2a}\ and\\x=\dfrac{-b-\sqrt D}{2a}[/tex]
Where [tex]D= b^{2} -4ac[/tex]
Comparing the equation with [tex]ax^{2} +bx+c=0[/tex]
a = 1
b = -4
c= -38
Calculating D,
[tex]D= (-4)^{2} -4(1)(-38)\\\Rightarrow D = 16+152 = 168[/tex]
Now, finding the roots:
[tex]x=\dfrac{-(-4)+\sqrt {168}}{2\times 1}\\\Rightarrow x=\dfrac{4+2\sqrt {42}}{2}\\\Rightarrow x=2+\sqrt {42}\\and\\x=\dfrac{-(-4)-\sqrt {168}}{2\times 1}\\\Rightarrow x=\dfrac{4-2\sqrt {42}}{2}\\\Rightarrow x=2-\sqrt {42}[/tex]
So, the solution is:
[tex]x=2$\pm$\sqrt{42}[/tex]
