How do you solve by using the quadratic formula: $x^2 + 7x = -3$ ?

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Answer 1 · 2016-05-09T19:22:19

$x=-7/2+-sqrt(37)/2$

First add $3$ to both sides to get:

$x^2+7x+3 = 0$

This is in the form $ax^2+bx+c = 0$ with $a=1$ , $b=7$ and $c=3$ .

The roots are given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$=(-7+-sqrt(7^2-(4*1*3)))/(2*1)$

$=(-7+-sqrt(49-12))/2$

$=(-7+-sqrt(37))/2$

$=-7/2+-sqrt(37)/2$

Note that since $37$ is prime, the square root does not simplify further.

$sqrt(37)$ does have a simple continued fraction expansion which you can truncate to give rational approximations:

$sqrt(37) = [6;bar(12)] = 6+1/(12+1/(12+1/(12+1/(12+1/(12+...)))))$

How do you solve by using the quadratic formula: x^2 + 7x = -3x^2 + 7x = -3?