How do you solve $x^{2} + 4 x + 4 = 7$ $x^2 + 4x + 4 = 7$ ?

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Answer 1 · 2015-07-27T16:28:58

The solutions are:
$x = - 2 + \sqrt{7}$ $color(green)(x=-2+sqrt(7)$
$x = - 2 - \sqrt{7}$ $color(green)( x=-2-sqrt(7)$

$x^{2} + 4 x + 4 = 7$ $x^2+4x+4=7$

$x^{2} + 4 x - 3 = 0$ $x^2+4x-3=0$

The equation is of the form $a x^{2} + b x + c = 0$ $color(blue)(ax^2+bx+c=0$ where:
$a = 1, b = 4, c = - 3$ $a=1, b=4, c=-3$

The Discriminant is given by:
$Delta=b^2-4*a*c$

$= (4)^2-(4*1*((-3))$

$= 16+12 =28$

The solutions are found using the formula
$x=(-b+-sqrtDelta)/(2*a)$

$x = ((-4)+-sqrt(28))/(2*1) = (-4+-2sqrt(7))/2$

$=(cancel(2)(-2+-sqrt(7)))/cancel2$

$color(green)(x=-2+sqrt(7), x=-2-sqrt(7)$

How do you solve x^2 + 4x + 4 = 7x2+4x+4=7x^2 + 4x + 4 = 7?