solve the quadratic equation by completing the square (plzz help!!)?

I'll solve by completing the square

x² + 3y = 10
x² + 3y +(3/2)² = 10 + (3/2)²
x² + 3y +(3/2)² = 10 + 9/4
(x + 3/2)² = 49/4
(x + 3/2) = ±7/2
x = –3/2 ±7/2
x = –3/2 –7/2 = –10/2 = –5
x = –3/2 +7/2 = 4/2 = 2
x = {–5, 2}

x² – 6x = 0
x² – 6x + 9 = 0 + 9
(x – 3) = ± 3
(x – 3) = ± 3
x = 3 ± 3
x = 3 – 3 = 0
x = 3 + 3 = 6
x = {0, 6}

2a² + 8a = 5 2(a² + 4a ) = 5 ...a million/2(4) then sq. this and upload on the two factors...on the applicable facet you may upload 2*4 because of the element 2 outdoors the () on the applicable facet 2(a² + 4a + 4) = 5 + 2*4 2(a + 2)² = thirteen the sq. is now achieved now sparkling up (a + 2)² = thirteen/2 a + 2 = ±?(thirteen/2) a = -2 ± ?(thirteen/2) which is additionally written as a = -2 ± ?(26) / 2 ~~

1)
y^2 + 3y = 10
y^2 + 3y/2 + 3y/2 = 10
y^2 + 3y/2 + 3y/2 + 9/4 = 10 + 9/4
(y^2 + 3y/2) + (3y/2 + 9/4) = 40/4 + 9/4
y(y + 3/2) + 3/2(y + 3/2) = 49/4
(y + 3/2)(y + 3/2) = 49/4
(y + 3/2)^2 = 49/4
y + 3/2 = ±√(49/4)
y = -3/2 ±7/2

y = -3/2 + 7/2
y = 4/2
y = 2

y = -3/2 - 7/2
y = -10/2
y = -5

∴ y = -5 , 2

= = = = = = = =

2)
x^2 - 6x = 0
x^2 - 3x - 3x = 0
x^2 - 3x - 3x + 9 = 0 + 9
(x^2 - 3x) - (3x - 9) = 9
x(x - 3) - 3(x - 3) = 9
(x - 3)(x - 3) = 9
(x - 3)^2 = 9
x - 3 = ±√9
x - 3 = ±3
x = 3 ±3

x = 3 + 3
x = 6

x = 3 - 3
x = 0

∴ x = 0 , 6

may i correct the guy!!

Question 1
y² + 3y - 10 = 0
(y² + 3y + 9/4) = 10 + 9/4
(x + 3/2)² = 49/4
x + 3/2 = ± 7/2
x = - 3/2 ± 7/2
x = - 5 or x = 2

Question 2
x² - 6x + 9 = 9
(x - 3) ² = 9
x - 3 = ± 3
x = 3 ± 3
x = 6 , x = 0

These are the true answers, sure !!!!

First is (y+5)(y-2) = 0
i.e. y = 2 or -5

Second is x^2 = 6x
so x = 6
gud luck
don't forget to give 10 pts

Question 1
y² + 3y - 10 = 0
(y² + 3y + 9/4) = 10 + 9/4
(x + 3/2)² = 49/4
x + 3/2 = ± 7/4
x = - 3/2 ± 7/4

Question 2
x² - 6x + 9 = 9
(x - 3) ² = 9
x - 3 = ± 3/2
x = 3 ± 3/2
x = 9/2 , x = 3/2

(y+5)(y-2) is the first