Question

 x² -3x-28=0 como resolver?

Answer 1

$ax^2+bx+c=0\\x^2-3x-28=0\\\\a=1\ \ \ \ \ \ \ b=-3\ \ \ \ \ \ \ c=-28\\\\\triangle=b^2-4ac\\\triangle=(-3)^2-4\cdot(1)\cdot(-28)\\\triangle=9+112\\\triangle=121$

$x= \dfrac{-b \frac{+}{-} \sqrt{\triangle}}{2a}\\\\\\x= \dfrac{-(-3) \frac{+}{-} \sqrt{121} }{2}\\\\\\x= \dfrac{3 \frac{+}{-}11}{2}\\\\\\x'= \dfrac{3+11}{2}\\\\\\x'= \dfrac{14}{2}\\\\\\x'=7\\\\\\x"= \dfrac{3-11}{2}\\\\\\x"= -\dfrac{8}{2}\\\\\\x"=-4$

$\boxed{S=\{(7,-4)\}}$