How do you find the solution to 2y-4(x)squared+6x+12=6?

Answer 1

Go find a really clever person - thats how!!

Answer 2

First, rearrange the equation to make y the subject. You will get y = 2x^2 - 3x - 3. This is a quadratic equation and solving means find the values of x for which y = 0. So you now need to solve 2x^2 - 3x -3 = 0. It won't factorise so you best use the quadratic formula. You can leave the answers in surd form or evaluate to get x ~ 2.186 or x ~ -0.986. These are the solutions to your equation. Of course there are many other x and y values that satisfy the equation but these are not solutions. If you chose a range of values for x and calculated y you could then draw a graph. It would be a parabola (U shaped) and cross the x axis at -0.986 and 2.186

Answer 3

2y - 4x^2 + 6x + 12 = 6

- 6 >> 2y - 4x^2 + 6x + 6 = 0

Transpose a bit (swap items over to make things easier):
add 4x^2 to the other side; subtract 6x from each side, thus giving you -6x on the right; and subtract 6, thus giving you -6 on the right-hand side.

2y = 4x^2 - 6x - 6


/2 >> y = 2x^2 - 3x - 3

This is a quadratic equation (see http://en.wikipedia.org/wiki/Quadratic_equation )

Quadratic: ax^2 + bx + c

a= 2 b = -3 c = -3

Use the quadratic formula on my link.

Answer 4

The solution is a curve plotted on an x / y Graph.

You can solve numerically only if you define one of the variables

eg. when X=0 , Y= -3 (so the line cuts the X axis once)

or for Y=0, when X will have 2 solutions (ie. the line cuts the Y axis twice)

Answer 5

2y - 4x2 +6x = -6

2 solutions are:
x=0, y=-3
y=2, x=-1 (by trial and error)

Answer 6

2y-4(x)squared+6x+12=6
y -2x^2 -3x +6 = 3
y = 2x^2 +3x -3
x= [-3+/- sqrt(9+24)]/ -4
x = 3/4 +/- (-.25sqrt(23)

Answer 7

Solve for each variable separately.

2y -4(x^2)+6x+12=6
2y-4(x^2)+6x+6=0
2y=4(x^2)-6x-6
y=2(x^2)-3x-3


2y -4(x^2)+6x+12=6
2y-4(x^2)+6x+6=0
2y=4(x^2)-6x-6
y=2(x^2)-3x-3
2(x^2)-3x=3-y
2(x^2)=3x+3-y
(x^2)=(1.5)x+1.5-y/2
(x^2)-(1.5)x=1.5-y/2
(x^2)/2=1.5+y/2
(x^2)=3+y
x=sqrt(3+y)

You can then substitute each into the other to solve for the number value.

Answer 8

You've got one equation with two unknowns, so it depends on which variable you'd like to have solved for, in terms of the other.

Answer 9

You can't, because there are two variables and only one equation.

Answer 10

dose it mater