What are the real roots of x2-7×+10=0

Is that x^2-7x+10? If so, use the AC method by multiplying the leading coefficient with the constant. Leading coefficient is 1 and constant is 10. 1x10 =10 and then you have to multiply two numbers to get 10 and add two number that give you -7. in this case it's -5 and -2. so that means x^2-5x-2x+10 and then factor out the first two x(x-5) and factor out the last two -2(x-5) combine the two (x-2)(x-5)=0 solve for both x values x = 2 and 5.