Question

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

Answer 1

Answer:

 $y_1=-\frac{1}{2}\\\\y_2=-\frac{3}{2}$

Step-by-step explanation:

The Quadratic formula is:

$y=\frac{-b\±\sqrt{b^2-4ac} }{2a}$

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:

$y=\frac{-8\±\sqrt{8^2-4(4)(3)} }{2(4)}$

$y_1=-\frac{1}{2}\\\\y_2=-\frac{3}{2}$

Answer 2

Answer:

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

Step-by-step 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

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

$y=\frac{-(8)+-\sqrt{(16)}}{8}$

$y=\frac{-8+4}{8}$ and $y=\frac{-8-4}{8}$

The solutions are

$y=\frac{-1}{2}$ and $y=\frac{-3}{2}$