Ok let’s solve it
5(x-2)^2-20=0
first let’s foil (x-2)
5(x^2-4x+4) -20=0
now distribute the 5
5x^2 -20x +20 -20 = 0
combine like terms
5x^2-20x=0
take the gcf
5x(x-4)=0
x=0, 4
solutions are (4,0) and (2, -20) because the original vertex form a(x-h)^2+k
50 + 3x = 7x + 12
This is because of the exterior angle theorem.
<span>Real world numbers are most likely to be compared in the settings wherein their functions are more effective and accurate than when they are not. Aspects such as accounting reasons and others are crucial in determining the most accurate results when it comes to these numbers.</span>
