What is the vertex of the quadratic y=6(x−2)2+10?

Given quadratic equation ,
${\displaystyle y=6{(x-2)}^2+10}$
On simplifying,
${\displaystyle y=6(x^2+4-4x)+10}$
${\displaystyle y=6x^2-24x+24+10}$ (1)
${\displaystyle y=6x^2-24x+34}$
Now, this equation compare with general quadratic equation,
${\displaystyle a{x}^2+bx+c}$
a = 6 , b = -24 , c = 34
${\displaystyle h=-\frac{b}{2}a}$
${\displaystyle =-\frac{-24}{2\cdot 6}}$
= 2
Now putting x= 2 in equation (i), then we get
${\displaystyle y=6\cdot 2^2-24\cdot 2+34}$
y = 24 - 48 + 34
y = 10
(2, 10) is the vertex of given quadratic equation.