How do you complete the square and solve $x^{2} + 3 x - 28 = 0$ $x^2+3x-28=0$ ?

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Answer 1 · 2016-11-28T22:31:00

$x = - 7$ $x =-7$ and $x = 4$ $x = 4$

First, to factor the quadratic equation we need to play with integers which multiple to 28: 1,28 and 2,14 and 4,7

Trying combinations of these allows us to rewrite this problem as:

$(x + 7) (x - 4) = 0$ $(x + 7)(x - 4) = 0$

We can now solve for $x + 7 = 0$ $x + 7 = 0$ and $x - 4 = 0$ $x - 4 = 0$

$x + 7 = 0 \to x + 7 - 7 = 0 - 7 \to x = - 7$ $x + 7 = 0 -> x + 7 - 7 = 0 - 7 -> x =-7$

$x - 4 = 0 \to x - 4 + 4 = 0 + 4 \to x = 4$ $x - 4 = 0 -> x -4 + 4 = 0 + 4 -> x = 4$

How do you complete the square and solve x2+3x−28=0x^2+3x-28=0?