Verified Answer :

nobillionaireNobley

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

calculista

Answer:

the x-intercepts are

\((-4,0)\) and \((5/2,0)\)

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form \(ax^{2} +bx+c=0\) is equal to

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

in this problem we have

\(f(x)=-2x^{2} -3x+20\)  

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

\(-2x^{2} -3x+20=0\)  

so

\(a=-2\\b=-3\\c=20\)

substitute in the formula

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

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

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

\(x=\frac{3+13} {-4}=-4\)

\(x=\frac{3-13} {-4}=5/2\)

therefore

the x-intercepts are

\((-4,0)\) and \((5/2,0)\)

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