How do you solve $x^{2} + 4 x - 13 = 5$ $x^ { 2} + 4x - 13= 5$ ?

Question

How do you solve $x^{2} + 4 x - 13 = 5$ $x^ { 2} + 4x - 13= 5$ ?

1 Answer

Impact of this question

1131 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-13T17:44:34

As follows

Explanation:

$x^{2} + 4 x - 13 = 5$ $x^2+4x-13=5$
$x^{2} + 4 x - 18 = 0$ $x^2+4x-18=0$
$x^{2} + 2 (2) x + {(2)}^{2} - {(2)}^{2} - 18 = 0$ $x^2+2(2)x+(2)^2-(2)^2-18=0$
${(x + 2)}^{2} - 4 - 18 = 0$ $(x+2)^2-4-18=0$
${(x + 2)}^{2} = 22$ $(x+2)^2=22$
$x + 2 = \pm \sqrt{22}$ $x+2=+-sqrt22$
$x = - 2 \pm \sqrt{22}$ $x=-2+-sqrt22$

Science

Math

Humanities

... and beyond

How do you solve $x^{2} + 4 x - 13 = 5$ $x^ { 2} + 4x - 13= 5$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve x2+4x−13=5x^ { 2} + 4x - 13= 5?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $x^{2} + 4 x - 13 = 5$ $x^ { 2} + 4x - 13= 5$ ?