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How do I use the quadratic formula to solve #5x^2+9x-2=0#?

Answer 1

Madelyn Anderson

#ax^2+bx+c=0#

#5x^2+9x-2=0#

#a=5, b=9, c=-2#

formula for quadratics

#x=(-(9)+-sqrt((9)^2-4(5)(-2)))/(2(5))#

#x=(-9+-sqrt(81+40))/10#

#x=(-9+-sqrt(121))/10#

#x=(-9+-11)/10#

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Answer 2

Anna Allen

To use the quadratic formula to solve (5x^2 + 9x - 2 = 0), follow these steps:

Identify the coefficients (a), (b), and (c) in the quadratic equation: (ax^2 + bx + c = 0).
In this equation, (a = 5), (b = 9), and (c = -2).
Substitute these values into the quadratic formula: (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}).
Plug in the values of (a), (b), and (c) into the quadratic formula.
Calculate the discriminant (b^2 - 4ac) and determine its value.
If the discriminant is positive, there are two real roots. If it's zero, there is one real root (a repeated root). If it's negative, there are no real roots (two complex roots).
Apply the quadratic formula to find the solutions for (x).
Express the solutions in simplified form.

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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.