How do you solve $3x^2 + 20x + 36 = 4$ by completing the square?

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Answer 1 · 2015-05-18T18:15:02

$f(x) = 3x^2 + 20x + 32$ = $3(x^2 + (20x)/3 + 32/3) = 0$

$x^2 + (20x)/3 + 100/9 - 100/9 + 32/3 = 0$

$= (x + 10/3)^2 - 4/9 = 0$ -> $(x + 10/3) = +- 2/3$

$(x + 10/3) = 2/3 -> x = -10/3 + 2/3 = -8/3$

$(x + 10/3) = -2/3 -> x = -10/3 - 2/3 = -12//3 = -4$

How do you solve 3x^2 + 20x + 36 = 43x^2 + 20x + 36 = 4 by completing the square?