Add a video to the Algebra 2 course for finding the roots of a quadratic function.

f(x)=6x^2-7x+3

f(x)=6(x^2-7/6x+3/6)

f(x)=6(x^2-7/6x+1/2)

p=-7/6

q=1/2

(-p/2+u)(-p/2-u)=q

(7/12+u)(7/12-u)=1/2

(7/12)^2-u^2=1/2

49/144 - 1/2 = u^2

49/144 - 72/144 = u^2

-23/144 = u^2

u=i sqrt(23)/12

Roots:  {7/12 + i sqrt(23)/12, 7/12 - i sqrt(23)/12}

f(x)=6x^2-7x+2

f(x)=6(x^2-7/6x+1/3)

p=-7/6, q=1/3

(7/12 - u)(7/12 + u)

49/144 - u^2 = 2/3

49/144 - 48/144 = u^2

u = sqrt(1/144) = 1/12

7/12 - 1/12, 7/12 + 1/12 = 6/12, 8/12

(x-6/12)(x-8/12)

x^2-6x/12-8x/12+48/144

x^2 - 14x/12 + 1/3

x^2 - 7/6x + 1/3

6(x^2-7/6x+1/3) = 6x^2-7x+2

Roots: {1/2, 1/3}