What is the answer to 1x^2-12x+5=(x-a)^2+b^2 ?

The answer to 1x^2-12x+5=(x-a)^2+b^2 is x=(a^2+b^2-5)/(2(-6+a));\quad a\ne 6

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1x^2-12x+5=(x-a)^2+b^2

1 x^{2} - 12 x + 5 = (x - a)^{2} + b^{2}

x = \frac{a ^{2} + b ^{2} - 5}{2 ( - 6 + a )}; a \neq = 6

Solution steps

1 \cdot x^{2} - 12 x + 5 = (x - a)^{2} + b^{2}

Multiply:

1 \cdot x^{2} = x^{2}

x^{2} - 12 x + 5 = (x - a)^{2} + b^{2}

Expand

(x - a)^{2} + b^{2} : x^{2} - 2 a x + a^{2} + b^{2}

x^{2} - 12 x + 5 = x^{2} - 2 a x + a^{2} + b^{2}

Move

5

to the right side

x^{2} - 12 x = x^{2} - 2 a x + a^{2} + b^{2} - 5

Subtract

x^{2} - 2 a x

from both sides

x^{2} - 12 x - (x^{2} - 2 a x) = x^{2} - 2 a x + a^{2} + b^{2} - 5 - (x^{2} - 2 a x)

Simplify

2 a x - 12 x = a^{2} + b^{2} - 5

Factor

2 a x - 12 x : 2 x (a - 6)

2 x (a - 6) = a^{2} + b^{2} - 5

Divide both sides by

2 (a - 6); a \neq = 6

x = \frac{a ^{2} + b ^{2} - 5}{2 ( - 6 + a )}; a \neq = 6

t, d e t a = d e t a^{- 1}

\frac{x + 3}{4} = \frac{( 5 x - 2 )}{3 + 1}

0.75 (6 x - 1) = 0.25 (8 x - 1)

35 x - 13 = 22 + 45 x

x, 428 \cdot \frac{x}{y} = 12.24

What is the answer to 1x^2-12x+5=(x-a)^2+b^2 ?
The answer to 1x^2-12x+5=(x-a)^2+b^2 is x=(a^2+b^2-5)/(2(-6+a));\quad a\ne 6