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

Question

asked Dec 7, 2020 126k views

2 Answers

Jonathon Oates · Answer 1 · 2020-12-08T18:32:38+0000

Answer:

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

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:

$y=(-8\±√(8^2-4(4)(3)) )/(2(4))$

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

Jeroen Bouman · Answer 2 · 2020-12-13T22:33:31+0000

Answer:

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)