Solve: \[{{x}^{2}}+7x-8=0\]

Hint: In this type of question we have to use the concept of solving a quadratic equation. We know that there are various methods of solving a quadratic equation such as completing the square method, using quadratic formulas and by splitting the middle term. Using the quadratic formula \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] gives us the roots of the equation easily.

Complete step by step answer:
Now, we have to solve the quadratic equation \[{{x}^{2}}+7x-8=0\].
Comparing the given quadratic equation with the standard quadratic equation \[a{{x}^{2}}+bx+c=0\] we get,
\[\Rightarrow a=1,b=7,c=-8\]
Now by using quadratic formula, we get roots of the quadratic equation \[a{{x}^{2}}+bx+c=0\] as:
\[\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
By substituting the values of a, b and c in the quadratic formula we can write,
\[\begin{align}
  & \Rightarrow x=\dfrac{-7\pm \sqrt{{{\left( 7 \right)}^{2}}-4\times 1\times \left( -8 \right)}}{2\times 1} \\
 & \Rightarrow x=\dfrac{-7\pm \sqrt{49+32}}{2} \\
 & \Rightarrow x=\dfrac{-7\pm \sqrt{81}}{2} \\
 & \Rightarrow x=\dfrac{-7\pm 9}{2} \\
 & \Rightarrow x=\dfrac{-7+9}{2}\text{ and }x=\dfrac{-7-9}{2} \\
 & \Rightarrow x=\dfrac{2}{2}\text{ and }x=\dfrac{-16}{2} \\
 & \Rightarrow x=1\text{ and }x=-8 \\
\end{align}\]

Hence, \[x=1\text{ and }x=-8\] are the roots of the equation \[{{x}^{2}}+7x-8=0\].

Note: In this type of question students have to note that the quadratic formula is used for solving a quadratic equation only if we compare the given equation with the standard form of quadratic equation i.e. \[a{{x}^{2}}+bx+c=0\]. Students have to note that quadratic formula is the easiest way and most efficient formula to find the roots of the quadratic equation. Also students have to note that the quadratic equation can also be solved by using splitting of the middle term and by completing the square. One of the students may solve the given quadratic equation by splitting of the middle term as follows:
\[\begin{align}
  & \Rightarrow {{x}^{2}}+7x-8=0 \\
 & \Rightarrow {{x}^{2}}+8x-x-8=0 \\
 & \Rightarrow x\left( x+8 \right)-1\left( x+8 \right)=0 \\
 & \Rightarrow \left( x+8 \right)\left( x-1 \right)=0 \\
 & \Rightarrow x=-8\text{ and }x=1 \\
\end{align}\]
Hence, \[x=1\text{ and }x=-8\] are the roots of the equation \[{{x}^{2}}+7x-8=0\].