Respuesta :

kudzordzifrancis

Answer:

[tex]x=5+\sqrt{31}[/tex] or [tex]x=5-\sqrt{31}[/tex]

Step-by-step explanation:

The given quadratic equation is

[tex]6=x^2-10x[/tex]

Rewrite in the form;

[tex]ax^2+bx+c=0[/tex]

This implies that;

[tex]x^2-10x-6=0[/tex]

This gives us a=1,b=-10,c=-6

We use the quadratic formula;

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

Plug in the values to get;

[tex]x=\frac{--10\pm \sqrt{(-10)^2-4(1)(-6)} }{2(1)}[/tex]

[tex]x=\frac{10\pm \sqrt{100+24} }{2}[/tex]

[tex]x=\frac{10\pm \sqrt{124} }{2}[/tex]

[tex]x=\frac{10\pm2\sqrt{31} }{2}[/tex]

[tex]x=5\pm\sqrt{31}[/tex]

[tex]x=5+\sqrt{31}[/tex] or [tex]x=5-\sqrt{31}[/tex]

carlosego

For this case we must find the solutions of the following quadratic equation:

[tex]6 = x ^ 2-10x\\x ^ 2-10x-6 = 0[/tex]

Where:

[tex]a = 1\\b = -10\\c = -6[/tex]

According to the quadratic equation we have:

[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}\\x = \frac {- (- 10) \pm \sqrt {(- 10) ^ 2-4 (1) (- 6)}} {2 (1)}\\x = \frac {10 \pm \sqrt {100 + 24}} {2}\\x = \frac {10 \pm \sqrt {124}} {2}[/tex]

[tex]x = \frac {10 \pm \sqrt {124}} {2}\\x = \frac {10 \pm \sqrt {31 * 4}} {2}[/tex]

[tex]x = \frac {10 \pm 2\sqrt {31}} {2}[/tex]

So, we have two roots:

[tex]x_ {1} = 5 + \sqrt {31}\\x_ {2} = 5- \sqrt {31}[/tex]

ANswer:

[tex]x_ {1} = 5 + \sqrt {31}\\x_ {2} = 5- \sqrt {31}[/tex]

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