Answer
5, 6 and 7
Work Step by Step
Assume that the three consecutive integers are $x,x+1$ and $x+2$.
According to the question,
${{x}^{2}}+\left( x+1 \right)\left( x+2 \right)=67$
Simplify the quadratic equation
$\begin{align}
& {{x}^{2}}+\left( x+1 \right)\left( x+2 \right)=67 \\
& {{x}^{2}}+{{x}^{2}}+2x+x+2=67 \\
& 2{{x}^{2}}+3x=67-2 \\
& 2{{x}^{2}}+3x-65=0
\end{align}$
The solution of the quadratic equation $a{{x}^{2}}+bx+c=0$ is given by, $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Compare $2{{x}^{2}}+3x-65=0$ with $a{{x}^{2}}+bx+c=0$
Here, $a=2,b=3$ and $c=-65$
Substitute the values of $a,b$ and $c$ in $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$\begin{align}
& x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
& =\frac{-3\pm \sqrt{{{3}^{2}}-4\cdot 2\cdot \left( -65 \right)}}{2\cdot 2} \\
& =\frac{-3\pm \sqrt{9+520}}{4} \\
& =\frac{-3\pm \sqrt{529}}{4}
\end{align}$
$\begin{align}
& x=\frac{-3\pm 23}{4} \\
& =\frac{-3+23}{4},\frac{-3-23}{4} \\
& =\frac{20}{4},\frac{-26}{4} \\
& =5,\frac{-13}{2}
\end{align}$
Here, we only consider the positive value, x=5.
Thus, the consecutive integers are 5, 6, and 7.
