without LaTex, we write x squared as x^2

you don't need the quadratic formula for this, but since you asked.x^2-5x+6 = 0

remember the quadratic formula.

for a quadratic ax^2 + bx + c = 0, x is given by:

x = [-b +/- sqrt(b^2 - 4ac)]/2a

x^2-5x+6 = 0

=> x = [5 +/- sqrt(25 - 4(6)]/2

=> x = [5 +/- sqrt(1)]/2

=> x = (5 + 1)/2 or (5 - 1)/2

so x = 3 , x = 2

x = [-1 +/- sqrt(1 - 4(-10))]/2x^2+x-10 = 0

=> x = [-1 +/- sqrt(41)]/2

=> x = [-1 + sqrt(41)]/2 or x = [-1 - sqrt(41)]/2

for the roots, we want x^2 - 2x - 8 = 0y=x^2-2x-8

=> (x - 4)(x + 2) = 0

=> x = 4 or x = -2 ...........these are the roots

this is the difference of two squaresy=x^2-4

we have

y = x^2 - 2^2

=> y = (x + 2)(x - 2)

for roots, we want

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

so x = -2 or x = 2

i'll get back to you on thisgraph y>or equal to x^2

i'll get back to you on this as wellgraph y<x^2-3