How do you solve $x^2 + 8x + 13 = 0$ ?

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Answer 1 · 2016-02-25T03:28:05

$x=-4+sqrt 3, -4-sqrt 3$

$x^2+8x+13=0$ is a quadratic equation in standard form $ax^2+bx+13$ , where $a=1, b=8, c=13$ .

The quadratic formula can be used to solve the equation.

$x=(-b+-sqrt(b^2-4ac))/(2a)$

Substitute the known values into the formula and solve for $x$ .

$x=(-8+-sqrt(8^2-(4*1*13)))/(2*1)$

Simplify.

$x=(-8+-sqrt(64-52))/2$

Simplify.

$x=(-8+-sqrt 12)/2$

Factor $sqrt 12$ .

$sqrt(2xx2xx3)=$

$sqrt(2^2xx3)=$

$2sqrt 3$

$x=(-8+-2sqrt3)/2$

Simplify.

$x=(cancel(-8)^-4+-cancel(2)^1sqrt 3)/cancel(2)^1$

Simplify.

$x=-4+-sqrt 3$

Solutions for $x$ .

$x=-4+sqrt 3$

$x=-4-sqrt 3$

How do you solve x^2 + 8x + 13 = 0x^2 + 8x + 13 = 0?