How do you solve $x^2 + 6x – 7 = 0$ using the quadratic formula?

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Answer 1 · 2016-04-14T08:28:01

$x=1$
$x=-7$

Given-

$x^2+7x-7=0$
$x^2+7x-x-7=0$
$x(x+7)-1(x+7)=0$
$(x-1)(x+7)$
$x-1=0$
$x=1$

$x+7=0$
$x=-7$

Answer 2 · 2016-04-14T08:30:14

The solutions are:

$x = 1$

$x = -7$

$x^2 + 6x - 7 =0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:

$a=1, b=6, c= - 7$

The Discriminant is given by:

$Delta=b^2-4*a*c$

$= (6)^2-(4* 1 * (- 7))$

$= 36 + 28 = 64$

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

$x = ((-6) +-sqrt(64))/(2*1) = (-6 +- 8)/2$

$x = (-6 + 8)/2 = 2 /2 = 1$

$x = (-6 - 8)/2 = -14 /2 = -7$

How do you solve x^2 + 6x – 7 = 0x^2 + 6x – 7 = 0 using the quadratic formula?