Respuesta :

We have the next equation

[tex]-11x=2x^2+15[/tex]

First we need o make the equation equal to zero

[tex]2x^2+11x+15=0[/tex]

we have a quadratic equation,we will have 2 solutions, so we will use the general formula to find the solutions to this equation

[tex]x_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]

in our case

a=2

b=11

c=15

[tex]x_{1,2}=\frac{-11\pm\sqrt[]{11^2-4(15)(2)}}{2(2)}_{}[/tex][tex]x_{1,2}=\frac{-11\pm\sqrt[]{121-120}}{4}=\frac{-11\pm1}{4}[/tex]

for the first solution

[tex]x=\frac{-11+1}{4}=-\frac{10}{4}=-\frac{5}{2}[/tex]

for the second solution

[tex]x=\frac{-11-1}{4}=\frac{-12}{4}=-3[/tex]

the solutions to the equation given is

[tex]x=-\frac{5}{2},\: x=-3[/tex]

the correct choices are B. and C.

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