Respuesta :

r3t40

If so, you can insert that y = -12 and solve for a which will equate to x.

[tex]-12=x^2-7x[/tex]

[tex]x^2-7x+12=0[/tex]

[tex](x-3)(x-4)=0\implies a_1=3,a_2=4[/tex]

So a could be 3 or 4 (has two solutions).

Hope this helps.

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