How do you solve $2(6 - x) = x(x + 5)$ using the quadratic formula?

Question

How do you solve $2(6 - x) = x(x + 5)$ using the quadratic formula?

2 Answers

Impact of this question

3269 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-06T14:39:22

$x=(-7+-sqrt(97))/2$

Explanation:

First let's operate to have the standard form:
$12-2x=x^2+5x$
$x^2+7x-12=0$

Using the quadratic formula,

$x=(-7+-sqrt(7^2-4*(-12)))/2$
$x=(-7+-sqrt(49+48))/2$
$x=(-7+-sqrt(97))/2$

Answer 2 · 2016-06-06T15:08:58

$x' = (-7 + sqrt(97))/2$

$x' = (-7- sqrt(97))/2$

Explanation:

$2(6-x)=x(x+5)$

$12-2x=x^2+5x$

$x^2+5x+2x-12=0$

$x^2+7x-12=0$

Your quadratic equation of the form

$ax^2+bx+c=0$

The discriminant is determined using the following formula:

$color(red)(Delta = b^2- 4ac)$

$a=1 \ ; \ b= 7 ; \ c=12$

$Delta = (7)^2 - 4xx1xx(-12)$

$Delta = 97$

Since $Delta>0 =>$ two real roots exist

$color(red)(x'= (-b + sqrt(Delta))/(2a)) " and " color(red)( x''= (-b - sqrt(Delta))/(2a))$

$x' = (-17 + sqrt(97))/2 ~=1.42$

$x''= (-17 - sqrt(97))/2~=-8.42$

Science

Math

Humanities

... and beyond

How do you solve $2(6 - x) = x(x + 5)$ using the quadratic formula?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 2(6 - x) = x(x + 5) 2(6 - x) = x(x + 5) using the quadratic formula?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you solve $2(6 - x) = x(x + 5)$ using the quadratic formula?