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

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Answer 1 · 2015-03-31T11:25:30

$-2x^2-3x-3 = 0$

Divide all terms of both sides of the equation by $(-2)$ so we are working with an expression that begins simply with $x^2$

$x^2 +3/2x + 3/2 = 0$

then move the constant $(-3/2)$ off the left-side of the equation by subtracting $3/2$ from both sides
$x^2+3/2x = -3/2$

To "complete the square" we need something of the form:
$(x+a)^2$
which equals $x^2+2ax +a^2$

Since we have computed the coefficient of $x$ to be $3/2$
$rarr a = 3/4$
and $a^2 = 9/4$

Add $9/4$ to both sides of the equation
$x^2+3/2x+9/4 = 9/4 - 3/2$

Rewrite the left-side as a square and simplify the right-side
$(x+3/2)^2 = 3/4$

Take the square root of both sides
$x+3/2 = +-sqrt(3)/2$

$x = -(3+sqrt(3))/2$
or
$x = -(3-sqrt(3))/2 = (sqrt(3)-3)/2$

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