How do you solve by completing the square: $x^2 + 3x - 18 = 0$ ?

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Answer 1 · 2015-03-30T14:13:23

In this way:

$x^2+3x-18=0rArrx^2+3x+(3/2)^2-(3/2)^2-18=0rArr$

$(x^2+3x+9/4)-9/4-18=0rArr(x+3/2)^2=18+9/4rArr$

$(x+3/2)^2=(72+9)/4rArr(x+3/2)^2=81/4rArr$

$x+3/2=+-9/2rArrx=-3/2+-9/2rArr$

$x_1=-3/2-9/2=-12/2=-6$

$x_2=-3/2+9/2=6/2=3$ .

How do you solve by completing the square: x^2 + 3x - 18 = 0x^2 + 3x - 18 = 0?