What are the solutions to te equation 2y2 plus 7y plus 3 equals 0?

Answer 1

Assuming the equation is supposed to be 2y2 + 7y + 3 = 0

The ys in this problem might be a bit off-putting but they are no different from xs in this case. They can be used and solved for in the same manor.

To solve quadratic equations there are quite a few options. The easiest is to factor. This equation factors into (y + 3)(2y + 1) because when these factors are multiplied together they equal 0, at least one of the factors must equal 0. You can set each of the factors to 0 and solve for y to find the solutions.

y + 3 = 0

y = -3 (subtract 3 from both sides)

2y + 1 = 0

2y = -1 (subtract 1 from both sides)

y = -1/2 (divide both sides by 2)

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

if you did not notice that this equation factors you can always use the quadratic formula. y = ( -b +- root(b2 - 4ac)) / 2a this formula assumes the equation is in the standard form ax2 + bx + c = 0

if we compare this standard form to our equation 2y2 + 7y + 3 = 0

we get the values a = 2 b = 7 and c = 3

if we plug these values into the quadratic equation we get

y = (-(7) +- root((7)2 - 4(2)(3))) / 2(2)

simplified

y = (-7 +- root(49 - 24)) / 4

y = (-7 +- root(25)) / 4

y = (-7 +- 5) / 4

because of the +- sign there are two possible solutions so we must account for both

y = (-7 + 5) / 4

y = -2 / 4

y = -1/2

y = (-7 - 5) / 4

y = -12 / 4

y = -3

the solutions are y = -1/2 and y = -3 this confirms our previous answers.