i posted two questions

[Steven G]

i learn to use quadratic when i have function \(\displaystyle \lambda^2\)

If someone actually told you to solve factored quadratic equations, then that person doesn't know what they're doing.
You should know then whenever you have (x-a)(x-b)=0, then the solutions are x=a and x=b.
I'm wondering how you would solve the following cubic equation (x-1)(x-3)(x-11)=0? Please post back your preferred method on solving such an equation.

 

[Steven G]

But you do have \(\displaystyle \lambda^2\) !!!

\(\displaystyle (6-\lambda)(7-\lambda) - 8 = 0\)

\(\displaystyle \lambda^2 - 13*\lambda + 34 = 0\)...............This is the quadratic equation that @Steven G was referring to.

Actually I wasn't. The OP tried solving \(\displaystyle (6-\lambda)(7-\lambda)=0\) by using the quadratic equation.
Fair enough, it was the wrong equation that s/he tried to solve but I just wanted (actually needed) to point out that using the quadratic formula when the quadratic is in factored form is ridiculous.