help on solving these algebra equations?

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1.

x² - 7x + 49/4 = 18 + 7/4

( x - 7/4 )² = 79/4

x - 7/4 = ± √79 / 2

x = 7/4 ± √79 /2

2.

y = (x² - 2x + 1) - 1 + 3

y = (x - 1)² + 2

1) x^2 - 7x = 18

x^2 - 7x + 49/4 = 18 + 49/4

(x - 7/2)^2 = 121/4

x - 7/2 = +/- 11/2

x = 7/2 +/- 11/2 = 18/2 or -4/2 = 9 or -2.

2) y = x^2 - 2x - 3

y + 3 = x^2 - 2x

y + 3 + 1 = x^2 - 2x + 1

y + 4 = (x - 1)^2

y = (x - 1)^2 - 4. Final.

1)

x² - 7x = 18

To complete the square, we want the left side to be in the form of (x² + bx). We already have this form, so we can go onto the next step.

To complete the square, start with x's coefficient (-7).

half it (-7/2)

square it (49/4)

add that to both sides:

x² - 7x + 49/4 = 18 + 49/4

Now we can factor the left side as a perfect square trinomial and simplify the right:

(x - 7/2)² = 72/4 + 49/4

(x - 7/2)² = 121/4

Square root of both sides:

x - 7/2 = ±11/2

Add 7/2 to both sides:

x = 7/2 ± 11/2

x = -4/2 and 18/2

x = -2 and 9

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2) Write in vertex form. This also requires a complete the square step:

y = x² - 2x - 3

We want the right side to be in the form described above, so add 3 to both sides:

y + 3 = x² - 2x

Now we can follow the steps as listed above to complete the square. Add 1 to both sides:

y + 4 = x² - 2x + 1

Factor the right side:

y + 4 = (x - 1)²

And solve for y again:

y = (x - 1)² - 4