First write it in vertex form :-
y= a(x - 2)^2 + 3 where a is some constant.
We can find the value of a by substituting the point (0.0) into the equation:-
0 = a((-2)^2 + 3
4a = -3
a = -3/4
so our equation becomes y = (-3/4)(x - 2)^2 + 3
Step-by-step explanation:

According to this trigonometric function, −C gives you the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:
![\displaystyle Phase\:[Horisontal]\:Shift → \frac{0}{\frac{1}{7}} = 0 \\ Period → \frac{2}{1}π = 2π](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Phase%5C%3A%5BHorisontal%5D%5C%3AShift%20%E2%86%92%20%5Cfrac%7B0%7D%7B%5Cfrac%7B1%7D%7B7%7D%7D%20%3D%200%20%5C%5C%20Period%20%E2%86%92%20%5Cfrac%7B2%7D%7B1%7D%CF%80%20%3D%202%CF%80)
Therefore we have our answer.
Extended Information on the trigonometric function
![\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{2}{B}π \\ Amplitude → |A|](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Vertical%5C%3AShift%20%E2%86%92%20D%20%5C%5C%20Phase%5C%3A%5BHorisontal%5D%5C%3AShift%20%E2%86%92%20%5Cfrac%7BC%7D%7BB%7D%20%5C%5C%20Period%20%E2%86%92%20%5Cfrac%7B2%7D%7BB%7D%CF%80%20%5C%5C%20Amplitude%20%E2%86%92%20%7CA%7C)
NOTE: Sometimes, your <em>vertical shift</em> might tell you to shift your graph below or above the <em>midline</em> where the amplitude is.
I am joyous to assist you anytime.
Hope this helps!! I just distributed when needed and solved for the unknown variable
Answer:
x=2, x=-2
Step-by-step explanation:
3x²-3+4x²-12=13
7x²=13+12+3
7x²=28
x²=4
x=-2
x=2