How do you solve $p^{2} - 3 p - 88 = 0$ $p^2-3p-88=0$ ?

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Answer 1 · 2016-08-22T09:05:31

$p = 11$ $p=11$

$p = - 8$ $p=-8$

$p^{2} - 3 p - 88 = 0$ $p^2-3p-88=0$

or

$p^{2} - 11 p + 8 p - 88 = 0$ $p^2-11p+8p-88=0$

or

$p (p - 11) + 8 (p - 11) = 0$ $p(p-11)+8(p-11)=0$

or

$(p - 11) (p + 8) = 0$ $(p-11)(p+8)=0$

or

$p - 11 = 0$ $p-11=0$

or

$p = 11$ $p=11$ ----------------------Ans $1$ $1$

or

$p + 8 = 0$ $p+8=0$

or

$p + 8 = 0$ $p+8=0$

or

$p = - 8$ $p=-8$ -------------------------Ans $2$ $2$

How do you solve p2−3p−88=0p^2-3p-88=0?