What is x if $- 4 {(x + 2)}^{2} + 3 x = - 5$ $-4(x+2)^2+3x=-5$ ?

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What is x if $- 4 {(x + 2)}^{2} + 3 x = - 5$ $-4(x+2)^2+3x=-5$ ?

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Answer 1 · 2016-01-11T15:09:04

$x = - \frac{9}{4}$ $x = -9/4$ or $x = - 1$ $x=-1$

Explanation:

First expand the expression and move the $- 5$ $-5$ to the left hand side, to get it into standard form
$- 4 (x^{2} + 4 x + 4) + 3 x + 5 = 0$ $-4(x^2+4x+4) +3x +5 = 0$
$- 4 x^{2} - 16 x - 16 + 3 x + 5 = 0$ $-4x^2 -16x -16 +3x +5 = 0$
$- 4 x^{2} - 13 x - 9 = 0$ $-4x^2 -13x -9 = 0$
$- (4 x^{2} + 13 x + 9) = 0$ $-(4x^2 +13x+9)=0$
$4$ $4$ and $9$ $9$ add to give $13$ $13$ , so the factors we need are $4 & 1$ $4&1$ and $9 & 1$ $9&1$
$- (4 x + 9) (x + 1) = 0$ $-(4x +9)(x+1) = 0$
$:. x = -9/4$ or $x=-1$

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What is x if $- 4 {(x + 2)}^{2} + 3 x = - 5$ $-4(x+2)^2+3x=-5$ ?

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What is x if -4(x+2)^2+3x=-5−4(x+2)2+3x=−5-4(x+2)^2+3x=-5?

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What is x if $- 4 {(x + 2)}^{2} + 3 x = - 5$ $-4(x+2)^2+3x=-5$ ?