Answer:
[tex]x = - 1 + \sqrt{17}\\and\\x = - 1 - \sqrt{17}\\[/tex]
Step-by-step explanation:
given equation
[tex]x^2 +2x +1 = 17[/tex]
subtracting 17 from both sides
[tex]x^2 +2x +1 = 17\\x^2 +2x +1 -17= 17-17\\x^2 +2x - 16 = 0\\[/tex]
the solution for quadratic equation
[tex]ax^2 + bx + c = 0[/tex] is given by
x = [tex]x = -b + \sqrt{b^2 - 4ac} /2a \\\\and \ \\-b - \sqrt{b^2 - 4ac} /2a[/tex]
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in our problem
a = 1
b = 2
c = -16
[tex]x =( -2 + \sqrt{2^2 - 4*1*-16}) /2*1 \\x =( -2 + \sqrt{4 + 64}) /2\\x =( -2 + \sqrt{68} )/2\\x = ( -2 + \sqrt{4*17} )/2\\x = ( -2 + 2\sqrt{17} )/2\\x = - 1 + \sqrt{17}\\and\\\\x = - 1 - \sqrt{17}\\[/tex]
thus value of x is
[tex]x = - 1 + \sqrt{17}\\and\\x = - 1 - \sqrt{17}\\[/tex]