Answer:
![\large\boxed{\red{\sf x =6+√(53),6-√(53)} }](https://img.qammunity.org/2023/formulas/mathematics/college/rcv4ugo0ngsmrfn1ty6ret5lr56jp6cyx0.png)
Explanation:
We would like to solve the given quadratic equation by completing the square method .The given equation is ;
![\longrightarrow x^2=12x +17](https://img.qammunity.org/2023/formulas/mathematics/college/709xhlyluw3cw9wjpvsitnj0gzs1n2tg14.png)
Subtract 12x to both sides ,
![\longrightarrow x^2-12x = 17](https://img.qammunity.org/2023/formulas/mathematics/college/6ftc5ngf3q0eymj1ydaqe9mjz89ergaebh.png)
Here the co-efficient of x² is already 1 . So , we shall rewrite it as ,
![\longrightarrow x^2-2(6)(x) = 17](https://img.qammunity.org/2023/formulas/mathematics/college/ih0xhco2bdqo9xu6s2j1sx7awyd5pbgmrv.png)
Add 6² on both sides ,
![\longrightarrow x^2-2(6)(x) + 6^2=17+6^2](https://img.qammunity.org/2023/formulas/mathematics/college/6kxgdbjg7fykfpynyeaiubbyimh4bh6eji.png)
Now LHS is in the form of (a-b)² = a²-2ab+b² ; so that ;
![\longrightarrow (x -6)^2 = 17+36](https://img.qammunity.org/2023/formulas/mathematics/college/u925cgkzk1syuodzigxv46bhl8osotojky.png)
Simplify RHS ,
![\longrightarrow (x-6)^2=53](https://img.qammunity.org/2023/formulas/mathematics/college/6oykerlmvyw5c95kw9nr5jlcbdqxs9ihqp.png)
Put square root on both sides,
![\longrightarrow (x-6) = \pm√(53)](https://img.qammunity.org/2023/formulas/mathematics/college/khfi5xsqwy4esdp7dfnyne242eq27h9j9c.png)
Add 6 on both sides,
![\longrightarrow x = 6\pm√(53)](https://img.qammunity.org/2023/formulas/mathematics/college/ocynki46ple3w3hat8ngnwsx9000n5wqjn.png)
Separate the two solutions ,
![\longrightarrow \underline{\underline{x =6+√(53),6-√(53)}}](https://img.qammunity.org/2023/formulas/mathematics/college/iq0j3kyqzlhv7vr62s1ui4pxnvjydoi6tc.png)
And we are done!