Question

how do you do 2x(x-3)=5x^2-7x????

Answer 1

Solve the equation and set equal to 0:
2x(x-3)=5x^2-7x
2x^2-6x=5x^2-7x
0 = 3x^2-x
0 = x(3x-1)

So set each part equal to 0, you get x=0 and x=1/3!

Answer 2

where did the other ^2 come from i been trying to get it and i was no where near that ansa??

Answer 3

When you distribute the 2x into the quantity (x-3)
You're doing:
2x(x) + 2x(-3) = 2x^2 - 6x

Answer 4

but i still didnt come up with that ansa i was trying to see where i went wrong...and thanks!!

Answer 5

Not a problem buddy!

Answer 6

i see i forgot to distribute the final part x(3x-1) smh thanks again