To solve: x^{2}+11x+24=0

Step 1
Given: ${\displaystyle x^2+11x+24=0}$
Split the middle term 11x as ${\displaystyle +3x}$ and ${\displaystyle +8x}$
Replace 11x with ${\displaystyle +3x+8x}$ in ${\displaystyle x^2+11x+24=0}$
${\displaystyle x^2+8x+3x+24=0}$
The above equation can be written ads ${\displaystyle x\times x+8\times x+3\times x+3\times 8}$
Apply distributive property for first two terms and last two terms,
${\displaystyle x\times x+8\times x+3\times x+3\times 8=x(x+8)+3(x+8)}$
Again apply distributive property for ${\displaystyle x(x+8)+3(x+8)}$
${\displaystyle x(x+8)+3(x+8)=(x+3)(x+8)}$
So the factors are ${\displaystyle (x+8)\text{and}(x+3)}$
${\displaystyle x^2+11x+24=(x+8)(x+3)}$
To find the value of x
Take ${\displaystyle x+8=0}$
Subtract 8 on both sides,
${\displaystyle x+8-8=0-8}$ NSk ${\displaystyle x=-8}$
Take ${\displaystyle x+3=0}$
Subtract 3 on both sides,
${\displaystyle x-3+3=0-3}$
${\displaystyle x=-3}$
So, the values of x are -8 and -3

Step 1
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form
${\displaystyle x^2+bx=c}$
${\displaystyle x^2+11x+24=0}$
Subtract 24 from both sides of the equation.
${\displaystyle x^2+11x+24-24=-24}$
Subtracting 24 from itself leaves 0
${\displaystyle x^2+11x=-24}$
Divide 11, the coefficient of the x term, by 2 to get ${\displaystyle \frac{11}{2}}$. Then add the square of ${\displaystyle \frac{11}{2}}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
${\displaystyle x^2+11x+{\left(\frac{11}{2}\right)}^2=-24+{\left(\frac{11}{2}\right)}^2}$
Square ${\displaystyle \frac{11}{2}}$ by squaring both the numerator and the denominator of the fraction.
${\displaystyle x^2+11x+\frac{121}{4}=-24+\frac{121}{4}}$
Add -24 to ${\displaystyle \frac{121}{4}}$
${\displaystyle x^2+11x+\frac{121}{4}=\frac{25}{4}}$
Factor ${\displaystyle x^2+11x+\frac{121}{4}}$. In general, when ${\displaystyle x^2+bx+c}$ is a perfect square, it can always be factored as ${\displaystyle {(x+\frac{b}{2})}^2}$
${\displaystyle {(x+\frac{11}{2})}^2=\frac{25}{4}}$
Take the square root of both sides of the equation.
${\displaystyle \sqrt{{(x+\frac{11}{2})}^2}=\sqrt{\frac{25}{4}}}$
Simplify.
${\displaystyle x+\frac{11}{2}=\frac{5}{2}}$
${\displaystyle x+\frac{11}{2}=-\frac{5}{2}}$
Subtract ${\displaystyle \frac{11}{2}}$ from both sides of the equation.
${\displaystyle x=-3}$
${\displaystyle x=-8}$

Step 1
Determine the quadratic equation's coefficients a, b and c
Use the standard form, $a{x}^2+bx+c=0$, to find the coefficients of our equation, $x^2+11x+24=0$
a=1
b=11
c=24
Step 2
Plug these coefficients into the quadratic formula
The quadratic formula gives us the roots for $a{x}^2+bx+c=0$, in which a, b and c are numbers (or coefficients), as follows:a=1
b=11
c=24

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\frac{-11\pm \sqrt{{11}^2-4\times 1\times 24}}{2\times 1}$
Simplify exponents and square roots
$\begin{array}{}x=\frac{-11\pm \sqrt{121-4\times 1\times 24}}{2\times 1}\\ \text{Perform any multiplication or division, from left to right:}\\ x=\frac{-11\pm \sqrt{121-4\times 24}}{2\times 1}\\ x=\frac{-11\pm \sqrt{121-96}}{2\times 1}\\ x=\frac{-11\pm \sqrt{25}}{2\times 1}\\ x=\frac{-11\pm \sqrt{25}}{2}\end{array}$
to get the result:
$x=\frac{-11\pm \sqrt{25}}{2}$
Step 3
Simplify square root $\sqrt{25}$
Simplify 25 by finding its prime factors
The prime factorization of 25 is $5^2$
Write the prime factors:
$\sqrt{25}=\sqrt{5\times 5}$
Group the prime factors into pairs and rewrite them in exponent form:
$\sqrt{5\times 5}=\sqrt{5^2}$
Use the rule $\sqrt{x^2}=x$ to simplify further:
$\sqrt{5^2}=5$
Step 4
Solve the equation for x
$x=\frac{-11\pm 5}{2}$
The $\pm$ means two answers are possible
Separate the equations: $x_1=\frac{-11+5}{2}$ and $x_2=\frac{-11-5}{2}$
$\begin{array}{}{x}_1=\frac{-11+5}{2}\\ x_1=\frac{-6}{2}\\ x_1=-3\\ x_2=\frac{-11-5}{2}\\ x_2=\frac{-16}{2}\\ x_2=-8\end{array}$