How do you solve d^ { 2} = 50- 23dd2=5023d?

2 Answers
Mar 14, 2017

See below

Explanation:

We have d^2=50-23dd2=5023d, or d^2+23d-50=0d2+23d50=0.
By factoring the latter, we arrive at (d+25)(d-2)=0(d+25)(d2)=0.
Thus, d=-25, 2d=25,2.

For more information on how to factor polynomials, click the link below.

http://www.instructables.com/id/How-to-factor/

d = 2 or -25d=2or25

Explanation:

This is a quadratic function.
First, you need to make it into a quadratic function which looks like this:

d^2 = 50 - 23dd2=5023d

Transpose

d^2 + 23d - 50 = 0d2+23d50=0

{(a = 1), (b = 23), (c = -50) :}a=1b=23c=50

ac = 1 * -50 = -50ac=150=50

Two factors of -5050 that give the result of bb, 2323 are:

-2 and 252and25

So

d^2 - 2d + 25d -50 = 0d22d+25d50=0

Factorize

d(d - 2) + 25 (d - 2) = 0d(d2)+25(d2)=0

(d-2) (d + 25) = 0(d2)(d+25)=0

d-2= 0d2=0

d= 0 + 2= 2d=0+2=2

or

d+25=0d+25=0

d= 0- 25= -25d=025=25

d= 2 or -25d=2or25