Respuesta :

angelicdragon12
we have to use the quadratic formula:

[tex]x= \frac{-b±\sqrt{b^{2}-4ac}}{2a} [/tex]

in your equation:
a = 4
b = -10
c = 5

[tex]x= \frac{10±\sqrt{(-10)^{2}-4(4)(5)}}{2(4)} [/tex]

we will do the positive part first:
[tex]x= \frac{10+ \sqrt{(-10)^{2}-4(4)(5)} }{2(4)} \\ x= \frac{10+ \sqrt{100-4(20)} }{8} \\ x= \frac{10+ \sqrt{100-80} }{8} \\ x= \frac{10+ \sqrt{20} }{8}[/tex]

[tex]x= \frac{10+2 \sqrt{5} }{8} \\ x= \frac{2(5+\sqrt{5})}{8}[/tex]

cancel common terms:
[tex]x = \frac{2(5+\sqrt{5})}{2(4)} \\ x= \frac{5+\sqrt{5}}{4} [/tex]

Now for the negative root:
[tex]x= \frac{10- \sqrt{(-10)^{2}-4(4)(5)} }{2(4)} \\ x= \frac{10- \sqrt{100-4(20)} }{8} \\ x= \frac{10- \sqrt{100-80} }{8} \\ x= \frac{10- \sqrt{20} }{8}[/tex]

[tex]x= \frac{10-2 \sqrt{5} }{8} \\ x= \frac{2(5-\sqrt{5})}{8} \\ x= \frac{5-\sqrt{5}}{4} [/tex]

The values of x are 1.81 or 0.68 i.e x = 1.81 or x = 0.69

The standard quadratic equation is given as [tex]ax^2+bx+c =0[/tex]

This question, we can easily write down the values of a, b and c

  • a = 4
  • b = -10
  • c = 5

The quadratic equation formula is given as

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

Let's substitute the values into the above equation and solve for x

[tex]x = \frac{-(-10)+-\sqrt{(-10)^2 - 4*4*5} }{2*4}\\x= \frac{10+-\sqrt{100-80} }{8}\\x = \frac{10+-\sqrt{20} }{8}\\x = \frac{10+4.47}{8}=1.81\\x = \frac{10-4.47}{8}=0.69[/tex]

From the above calculations, the values of x are 1.81 or 0.69

x = 1.81 or x = 0.69

Learn more on quadratic equation here;

https://brainly.com/question/7784687

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