How do I solve for x in these quadratic equation?

Question #1

2(x-5)^2 = 12

2(x^2 - 10x + 25) = 12

x^2 - 10x + 25 = 6

x^2 - 10x + 19 = 0

I don't know how to continue with this one without using the quadratic formula...

Question #2

3(x+1)^2 = 10

3(x^2 + 2x + 1) = 10

3x^2 + 6x - 9 = 0

3(x^2 + 2x - 3) = 0

3(x + 3)(x - 1) = 0

x = -3, 1

Hope this helps, and good luck!! :)

2(x-5)^2=12

2[x^2-10x +25]=12

2 x^2-20x+50=12

2 x^2-20x+38=0

On factorization U will get decimal roots….

Another approach is:

(x-5)^2=6

(x-5)=√6 or -√6

X=(5+√6) or (5 -√6)(on simplifying these numbers will be decimals)

2)3(x+1)^2=10

Similar second approach:

(x+1)^2=(10/3)

(x+1)=√ (10/3) or -√(10/3)

X= -1-√ (10/3) or-1 -√(10/3)

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