How do you solve the quadratic using the quadratic formula given \[{{k}^{2}}+5k-6=0\] ?

Hint: Now we are given with the equation of the form $a{{x}^{2}}+bx+c$ where the variable is k. Now we know that for any equation in this form the roots of the quadratic equation is given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Hence using this formula we get the required roots.

Complete step-by-step solution:
Now we are given with a quadratic equation in k. We want to find the solution of the equation. Hence we want to find the value of k such that it satisfies the equation.
Now we know that the general form of a quadratic equation is $a{{x}^{2}}+bx+c=0$ .
Now comparing the given equation with the general equation we get the variable is k and a = 1, b = 5 and c = - 6.
Now for any quadratic equation of the form $a{{x}^{2}}+bx+c=0$ the roots of the equation are given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Now substituting the values of a, b and c we get,
\[\begin{align}
  & \Rightarrow x=\dfrac{-5\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( -6 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{-5\pm \sqrt{25+24}}{2} \\
 & \Rightarrow x=\dfrac{-5\pm \sqrt{49}}{2} \\
\end{align}\]
\[x=\dfrac{-5+7}{2}\] or \[x=\dfrac{-5-7}{2}\]
Hence the value of x is 1 or – 6.
Hence the roots of the equation are x = 1 and x = - 6.

Note: Now note that a quadratic equation always represents a parabolic curve. Now roots are nothing but the intersection of the x axis and the curve. Now we have three possibilities, either the curve cuts the parabola two times in which we get two distinct roots and the Discriminant D > 0. Where $D={{b}^{2}}-4ac$ .
Now if the curve touches the parabola exactly once then we have there are repeating roots and the discriminant D = 0. Now if the curve doesn’t touch the parabola at all then there are no real solutions and in this case we have D < 0.