Solve the following quadratic equation for all values of x in simplest form. 4(3x - 5)^2+5 = 13 (will give

Step-by-step explanation:

4(3x-5)^2+5=13

4(3x-5)^2+5-13=0

As  (a-b)^2=a^2-2ab+b^2

4((3x)^2-2(3x)(5)+(5)^2)-8=0

4(9x^2-30x+25)-8=0

36x^2-120x+100-8=0

36x^2-120x+92=0

using the Quadratic Formula where

a = 36, b = -120, and c = 92

x=(−bc± sqrt(b^2-4ac))/2a

x=(-(-120)±sqrt((−120)^2−4(36)(92))/2(36)

=120±sqrt(14400−13248−)/72

=120±sqrt(1152)/72

The discriminant b2−4ac>0

so, there are two real roots.

x=(120±24sqrt(2))/72

x=(120/72)±24sqrt(2)/72

implify fractions and/or signs:

x=5/3±sqrt(2)/3

hich becomes

x=2.13807

x=1.19526

a

step-by-step explanation:

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~