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

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Answer 1 · 2017-03-14T02:49:38

See below

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

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

Answer 2 · 2017-03-14T03:03:17

#d = 2 or -25#

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

#d^2 = 50 - 23d#

Transpose

#d^2 + 23d - 50 = 0#

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

#ac = 1 * -50 = -50#

Two factors of #-50# that give the result of #b#, #23# are:

#-2 and 25#

So

#d^2 - 2d + 25d -50 = 0#

Factorize

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

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

#d-2= 0#

#d= 0 + 2= 2#

or

#d+25=0#

#d= 0- 25= -25#

#d= 2 or -25#