Answer:
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
