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## Answered question

2021-09-11

Find the solutions of the equation.
${x}^{2}+4x+13=0$

### Answer & Explanation

lobeflepnoumni

Skilled2021-09-12Added 99 answers

Step 1
The given equation is:
${x}^{2}+4x+13=0$...(1)
As, the given equation is a quadratic equation.
The general form f a quadratic equation is:
$a{x}^{2}+bx+c=0$...(2)
So, finding the roots of any quadratic equation, we use:
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
On comparing equation (1) with the equation(2), we have:
a=1
b=4
c=13
Step 2
Solving it, we get:
$x=\frac{-4±\sqrt{{\left(-4\right)}^{2}-4\left(1\right)\left(13\right)}}{2\left(1\right)}$
$x=\frac{-4±\sqrt{16-52}}{2}$
$x=\frac{-4±\sqrt{-36}}{2}$
$x=-2±\sqrt{-9}$
$x=-2±3\sqrt{-1}$
Since, $\sqrt{-1}=i$
So, $x=-2±3i$
Hence, the quadratic equation have complex roots.
Therefore, we have:
x=-2+3i
x=-2-3i
The required solution of the given quadratic equation.

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