Respuesta :

Answer:

5 or -7

Step-by-step explanation:

Use the quadratic formula to solve for x:

[tex]x^2+2x-35=0\\x=\frac{-b\±\sqrt{b^2-4ac} }{2a} \\x=\frac{-2\±\sqrt{2^2-4(1)(-35)} }{2(1)} \\x=\frac{-2\±\sqrt{4+140} }{2}\\x=\frac{-2\±\sqrt{144} }{2}\\x=\frac{-2\±12}{2}\\x=-1\±6[/tex]

Simplify to get two real solutions, 5 and -7.

[tex]\\ \sf\longmapsto x^2+2x-35=0[/tex]

[tex]\\ \sf\longmapsto x^2-7x+5x-35=0[/tex]

[tex]\\ \sf\longmapsto x(x-7)+5(x-7)=0[/tex]

[tex]\\ \sf\longmapsto (x+5)(x-7)=0[/tex]

[tex]\\ \sf\longmapsto x=-5\:or\:x=7[/tex]

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