How do you solve $5x^2-40x+80=0$ ?

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Answer 1 · 2016-05-12T00:27:37

I think the answer is -5.3.

$5x^2-40x+80 =0$

$5x*5x=25x$ Take care of the exponent first.

$25x-40x+80=0$ Solve the like terms.

$-15x+80=0$ Carry 80 over.

$-15x/-15=80/-15$ Cancel out -15 and divide 80 by that same number.

$x=-5.3$ Plug in the answer you get and round it to the nearest ten.

How do you solve 5x^2-40x+80=05x^2-40x+80=0?