kaliathomas080 kaliathomas080

09-12-2020
Mathematics

contestada

What is the solution set for this quadratic equation 2(x-3)^2-4=28

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konnievas konnievas

09-12-2020

4(x-3)-4=28
4x-12-4=28
4x-16=28
4x=28+16
4x=44
x=11

Answer Link

09-12-2020

Answer: x₁=7, x₂=-1

Step-by-step explanation:

(x-3)^2= x^2-6x+9

2(x^2-6x+9)=28+4

x^2-6x+9=16

x^2-6x-7=0

you use the quadratic equation will give you 2 roots,

x₁=7, x₂=-1

Answer Link

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