Respuesta :

nobillionaireNobley
[tex]x= \frac{3 \pm \sqrt{ (-3)^{2} -4 \times -2 \times 20} }{2 \times -2} \\ =\frac{3 \pm \sqrt{ 9 + 160} }{-4} \\ =\frac{3 \pm \sqrt{ 169} }{-4} \\ = \frac{3+13}{-4} \ or \ \frac{3-13}{-4} \\ =-4 \ or \ 2.5[/tex]
x-intercepts is (-4, 0) and (5/2, 0)
calculista

Answer:

the x-intercepts are

[tex](-4,0)[/tex] and [tex](5/2,0)[/tex]

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to

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

in this problem we have

[tex]f(x)=-2x^{2} -3x+20[/tex]  

Equate the quadratic equation to zero to find the x-intercepts

[tex]-2x^{2} -3x+20=0[/tex]  

so

[tex]a=-2\\b=-3\\c=20[/tex]

substitute in the formula

[tex]x=\frac{3(+/-)\sqrt{(-3)^{2}-4(-2)(20)}} {2(-2)}[/tex]

[tex]x=\frac{3(+/-)\sqrt{9+160}} {-4}[/tex]

[tex]x=\frac{3(+/-)13} {-4}[/tex]

[tex]x=\frac{3+13} {-4}=-4[/tex]

[tex]x=\frac{3-13} {-4}=5/2[/tex]

therefore

the x-intercepts are

[tex](-4,0)[/tex] and [tex](5/2,0)[/tex]

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