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Solve for xxx. Write the smaller solution first, and the larger solution second. 2x^2 - 2x - 180 = 02x 2 −2x−180=02, x, squared, minus, 2, x, minus, 180, equals, 0 \text{smaller }x =smaller x=start text, s, m, a, l, l, e, r, space, end text, x, equals \text{larger } x =larger x=start text, l, a, r, g, e, r, space, end text, x, equals

Respuesta :

Kentung12345

Answer:

lesser x: -10.65

greater x: -5.35

facundo3141592

The two solutions for the quadratic equation are:

x = -9 (smaller) and x = 10 (the larger one).

How to solve the quadratic equation?

Here we have the quadratic equation:

2x^2 - 2x - 180 = 0.

The solutions of this equation are given by Bhaskara's formula, it is:

[tex]x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4*2*(-180)} }{2*2} \\\\x = \frac{2 \pm 38 }{4}[/tex]

So there are two solutions, one for each sign of the square root.

The smaller solution is:

x = (2 - 38)/4 = -9

The larger solution is:

x = (2 + 38)/4 = 10

If you want to learn more about quadratic equations, you can read:

https://brainly.com/question/1214333

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