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

Answer 1

Genesis Allen

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.

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Answer 2

Savannah Anderson

To solve the equation (5x^2 - 40x + 80 = 0), you can use the quadratic formula: (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a = 5), (b = -40), and (c = 80). Substituting these values into the quadratic formula, you get (x = \frac{{40 \pm \sqrt{{(-40)^2 - 4(5)(80)}}}}{{2(5)}}). Simplifying this expression yields two roots: (x = 4 + 4i) and (x = 4 - 4i), where (i) represents the imaginary unit. Therefore, the solutions to the equation are (x = 4 + 4i) and (x = 4 - 4i).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.