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Use the quadratic formula to solve. Round to the nearest tenth if necessary.
2x^2 -3x -7 = 0

Sagot :

reyloforlife

The quadratic formula is:

[tex]\frac{-b + \sqrt{b^2 - 4ac} }{2a}[/tex] and  [tex]\frac{-b - \sqrt{b^2 - 4ac} }{2a}[/tex]

In this case, a = 2, b = -3, and c = -7

So, we can plug in the numbers to get:

  [tex]\frac{-(-3) + \sqrt{(-3)^2 - 4(2)(-7)} }{2(2)}[/tex] and    [tex]\frac{-(-3) - \sqrt{(-3)^2 - 4(2)(-7)} }{2(2)}[/tex]

Simplifying, we get:

    [tex]\frac{3 + \sqrt{65} }{4}[/tex] and   [tex]\frac{3 - \sqrt{65} }{4}[/tex]

You need to use a calculator to find what these would be in decimal form. The answer, rounded to the nearest tenth, is: x = 2.8, x = -1.3

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