Question

Use the quadratic formula to solve 4y^2+ 8y +7 =4

Answer 1

`y_1=-(1)/(2)\\\\y_2=-(3)/(2)`

Explanation:

The Quadratic formula is:

`y=(-b\±√(b^2-4ac) )/(2a)`

Given the equation
`4y^2+ 8y +7 =4`, you need to subtract 4 from both sides:

`4y^2+ 8y +7 -4=4-4`

`4y^2+ 8y +3 =0`

Now you can identify that:

`a=4\\b=8\\c=3`

Then you can substitute these values into the Quadratic formula. Therefore, you get these solutions:

`y=(-8\±√(8^2-4(4)(3)) )/(2(4))`

`y_1=-(1)/(2)\\\\y_2=-(3)/(2)`

Answer 2

The solutions are y=-1/2 and y=-3/2

Explanation:

Ok, for this problem we need to use the quadratic formula:

For
`ax^(2) +bx+c=0`

The values of x which are the solutions of the equation:

`x=\frac{-b+-\sqrt{b^(2)-4ac}}{2a}`

In this case your variable is y, so:

`ay^(2) +by+c=0`

`y=\frac{-b+-\sqrt{b^(2)-4ac}}{2a}`

So, a=4, b=8 and c=3

`y=\frac{-(8)+-\sqrt{(8)^(2)-4(4)(3) } }{2(4)}`

`y=(-(8)+-√((16)))/(8)`

`y=(-8+4)/(8)` and
`y=(-8-4)/(8)`

The solutions are

`y=(-1)/(2)` and
`y=(-3)/(2)`