Respuesta :

luisejr77

Answer:

[tex]x_1=4.158[/tex]

[tex]x_2=0.8416[/tex]

Step-by-step explanation:

For an equation of the form [tex]ax^2 +bx +c[/tex]

The quadratic formula is

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

In this case the equation is:

[tex]2x^2 - 10x + 7 = 0[/tex]

Then

[tex]a= 2\\b= -10\\c= 7[/tex]

Therefore, using the quadratic formula we have:

[tex]x=\frac{-(-10)\±\sqrt{(-10)^2 -4(2(7)}}{2(2)}[/tex]

[tex]x=\frac{5\±\sqrt{11}}{2}[/tex]

[tex]x_1=4.158[/tex]

[tex]x_2=0.8416[/tex]

kudzordzifrancis

ANSWER

[tex]x = \frac{ 5 }{2} - \frac{\sqrt {11} } {2} \: or \: x = \frac{ 5 }{2} + \frac{\sqrt {11} } {2} [/tex]

EXPLANATION

The given quadratic equation is:

[tex] 2{x}^{2} - 10x + 7 = 0[/tex]

Comparing this equation to

[tex] a{x}^{2} + bx + c = 0[/tex]

we have

a=2, b=-10 and c=7.

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 get,

[tex]x = \frac{ - - 10 \pm \sqrt{ {( - 10)}^{2} - 4(2)(7) } }{2(2)} [/tex]

This implies that

[tex]x = \frac{ 10 \pm \sqrt{ {100} - 56 } }{4} [/tex]

[tex]x = \frac{ 10 \pm \sqrt {44 } }{4} [/tex]

[tex]x = \frac{ 10 \pm 2\sqrt {11 } }{4} [/tex]

[tex]x = \frac{ 5 \pm \sqrt {11 } }{2}[/tex]

[tex]x = \frac{ 5 }{2} - \frac{\sqrt {11} } {2} \: or \: x = \frac{ 5 }{2} + \frac{\sqrt {11} } {2} [/tex]

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