Finding the roots of a quadratic equation?

Question

Finding the roots of a quadratic equation with leading coefficient greater than 1:

4y^2=2y+6

(If there is more than one solution, separate them with commas.)

Answer 1

4y^2 - 2y - 6 = 0
2(2y^2 - y - 3) = 0
2(2y - 3)(y + 1) = 0

2y - 3 = 0
2y = 3
y = 3/2

y + 1 = 0
y = -1

y = 3/2, -1

Answer 2

4y^2 = 2y + 6

Move everything to the left hand side.

4y^2 - 2y - 6 = 0

Divide each coefficient by 2.

2y^2 - y - 3 = 0

Factor.

(2y - 3) (y + 1) = 0

Equate each factor to 0 to get the solutions.

2y - 3 = 0, 2y = 3, y = 3/2
y + 1 = 0, y = -1

Therefore,

y = {3/2, -1}

Answer 3

Given equation is x^2 - 11x + 28 = 0 Now evaluate the aspects of the consistent fee such that as quickly as they're extra or subtracted we get 11 because of the fact the linked fee so 7 * 4 = 28 and seven + 4 = 11 may well be seen be careful on occasion we'd get 2 or greater pairs like an identical element then we'd desire to apply all of them and verify which one ends up in answer now x^2 -4x -7x +28=0 (right here cut up 11 in terms of the aspects) take x as basic from 1st area and seven as basic from 2d area x(x-4)-7(x-4)=0 right here we'd desire to confirm that (x-4) might desire to be modern-day in the two aspects (x-7)(x-4)=0 now equate the two equations seperately to 0 x-7=0 x-4=0 x=7 x=4 so 7 and four are the roots of quadratic equation the above defined technique is one technique we've one technique it particularly is that if the equation is in variety ax^2 + bx + c = 0 Then the roots may well be found out making use of the formulation x = (-b + (b^2 -4ac)^a million/2 )/2a or (-b - (b^2 -4ac)^a million/2 )/2a so right here in given equation we've a=a million b=-11 c=28 now if we be conscious interior the above formulation x=(11+(11^2-(28*4*a million))^a million/2))/2*a million =(11+(121-112)^a million/2))/2 =(11+(9)^a million/2)/2 =(11+3)/2 =7 x=(11-(11^2-(28*4*a million))^a million/2))/2*a million =(11-(121-112)^a million/2))/2 =(11-(9)^a million/2)/2 =(11-3)/2 =4 so roots are 7 and four now did u get it! bye

Answer 4

4y^2=2y+6
4y^2 - 2y - 6 = 0
Divide by 2:
2y^2 - y - 3 = 0
(2y - 3)(y + 1) = 0
y = 3/2 or y = -1.

Answer 5

4y² - 2y - 6 = 0
2y² - y - 3 = 0
(2y - 3) (y + 1) = 0
y = 3/2 , y = - 1

Answer 6

4y^2-2y-6=0
divide by 2
2y^2-y-3=0
2y^2-3y+2y-3=0
y(2y-3+1(2y-3)=0
(2y-3)(y+1)=0
y=-1,3/2ANS