How do you solve $x^2 - 14x + 49 = 3$ ?

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Answer 1 · 2015-10-30T05:02:14

$x= 7±sqrt(3)$

You make the equation equal zero so you subtract three from both sides.

$x^2 -14x+46 = 0$
You can't factor it so you can either complete the square or use the quadratic formula

$x=(-b±sqrt(b^2-4ac))/(2a)$

$a=1, b=-14, c=46$
Insert that and you get

$x=(-(-14)±sqrt((-14)^2-4(1)(46)))/(2(1))$

$x= (14±sqrt(196-184))/(2)$

$x= (14±sqrt(12))/(2)$

Factor $sqrt(12)$ to $sqrt(4*3$ then square root the 4 to 2 making $2sqrt(3)$

$x= (14±2sqrt(3))/(2)$
Simplify

$x= 7±sqrt(3)$

How do you solve x^2 - 14x + 49 = 3x^2 - 14x + 49 = 3?