The solution of the given quadratic equations would be [tex]x = 6 \pm 3\sqrt{2}i[/tex].
How to find the solution to a standard quadratic equation?
Suppose the given quadratic equation is
[tex]ax^2 + bx + c = 0[/tex]
Then its solutions are given as
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
We have given a quadratic equation;
[tex]x^2-12x+36=90\\\\x^2-12x - 54 = 0[/tex]
The solution ;
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
a = 1
b = -12
c = -54
[tex]x = \dfrac{12 \pm \sqrt{(-12)^2 - 4\times 1\times (-54)}}{2}\\\\\\x = \dfrac{12 \pm \sqrt{144 - 216}}{2}\\\\\\x = \dfrac{12 \pm \sqrt{72}}{2}\\\\\\x = 6 \pm 3\sqrt{2}i[/tex]
Hence, the solution of the given quadratic equations would be [tex]x = 6 \pm 3\sqrt{2}i[/tex].
Learn more about quadratic equations here:
https://brainly.com/question/3358603
#SPJ5
