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chikafujiwara

Answer:

x=(5+i*sqrt(35))/2, (5-i*sqrt(35))/2

Step-by-step explanation:

Use the quadratic formula with the following values.

a = 1

b = -5

c = 15

Substitute and simplify.

(5+-sqrt((-5)^2-4*(1*15)))/2*1

x = (5+-i*sqrt(35))/2

When you get a negative number inside the square root, remember that you can pull out i to make the number inside positive.

kudzordzifrancis
ANSWER

[tex]x = \frac{5}{2} - \frac{ \sqrt{35}i }{2} \: or \: x = \frac{5}{2} + \frac{ \sqrt{35}i }{2} [/tex]

EXPLANATION

The given equation is

[tex]y = {x}^{2} - 5x + 15[/tex]

To solve this equation, we equate it to zero.

[tex]{x}^{2} - 5x + 15 = 0[/tex]

Comparing to ax²+bx+c=0, we have a=1, b=-5, c=15.

The solution is given by the quadratic formula,

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

We plug in the values to obtain,

[tex]x = \frac{ - - 5\pm \sqrt{ {( - 5)}^{2} - 4(1)(15) } }{2(1)} [/tex]

[tex]x = \frac{5\pm \sqrt{ 25 -60 }}{2} [/tex]

[tex]x = \frac{5\pm \sqrt{ - 35}}{2} [/tex]

Recall that,

[tex]\sqrt{-1}=i[/tex]

[tex]x = \frac{5\pm \sqrt{ 35}i}{2} [/tex]

[tex]x = \frac{5}{2} - \frac{ \sqrt{35}i }{2} \: or \: x = \frac{5}{2} + \frac{ \sqrt{35}i }{2} [/tex]

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