You can try finding the roots of the given quadratic equation to get to the solution of the equation.
There are two solutions to the given quadratic equation
[tex]x = 0.202, x = 113.798[/tex]
Suppose that the given quadratic equation is [tex]ax^2 + bx +c = 0[/tex]
Then its roots are given as:
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
First we will convert it in the aforesaid standard form.
[tex]102x + 11 = (x-6)^2 - 2\\102x + 11 + 2 = x^2 + 36 - 12x\\0 = x^2 -114x + 23\\x^2 -114x + 23 = 0\\[/tex]
Thus, we have
a = 1. b = -114, c = 23
Using the formula for getting the roots of a quadratic equation,
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\x = \dfrac{114 \pm \sqrt{114^2 - 92}}{2} \\\\ x = 0.202 (\text{used "-" sign})\\\\x = 113.798 ( used "+" sign})[/tex]
Thus, there are two solutions to the given quadratic equation
[tex]x = 0.202, x = 113.798[/tex]
Learn more here about quadratic equations here:
https://brainly.com/question/3358603
