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

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

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Answer 1 · 2016-02-25T03:28:05

$x = - 4 + \sqrt{3}, - 4 - \sqrt{3}$ $x=-4+sqrt 3, -4-sqrt 3$

Explanation:

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

The quadratic formula can be used to solve the equation.

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

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

$x = \frac{- 8 \pm \sqrt{8^{2} - (4 \cdot 1 \cdot 13)}}{2 \cdot 1}$ $x=(-8+-sqrt(8^2-(4*1*13)))/(2*1)$

Simplify.

$x = \frac{- 8 \pm \sqrt{64 - 52}}{2}$ $x=(-8+-sqrt(64-52))/2$

Simplify.

$x = \frac{- 8 \pm \sqrt{12}}{2}$ $x=(-8+-sqrt 12)/2$

Factor $\sqrt{12}$ $sqrt 12$ .

$\sqrt{2 \times 2 \times 3} =$ $sqrt(2xx2xx3)=$

$\sqrt{2^{2} \times 3} =$ $sqrt(2^2xx3)=$

$2 \sqrt{3}$ $2sqrt 3$

$x = \frac{- 8 \pm 2 \sqrt{3}}{2}$ $x=(-8+-2sqrt3)/2$

Simplify.

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

Simplify.

$x=-4+-sqrt 3$

Solutions for $x$ .

$x=-4+sqrt 3$

$x=-4-sqrt 3$

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

1 Answer

Explanation:

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