Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

How do you solve these quadratic equations? Listed below. ?

- 12d^2 + 2d - 30 

- 4y^2 + 8y + 4 

6 Answers

Relevance
  • 1 month ago

    You cannot solve these as the variables are different in each expression:

    one is "d" and the other is "Y".

    They can Be factored:

    d = 1/12 (1 - i sqrt(359))d = 1/12 (1 + i sqrt(359))

    4 (-y + sqrt(2) + 1) (y + sqrt(2) - 1)

  • 1 month ago

    Mistake: what you shew are not equations but

    algebraic expressions.

    You may use the quadratic formula:

    x=[-b+/-sqr(b^2-4ac)]/(2a)

    where ax^2+bx+c=0 is the quadratic equation.

  • David
    Lv 7
    1 month ago

    Without an equality sign that equals something the given expressions are not equations.

  • rotchm
    Lv 7
    1 month ago

    Hint: Ever heard/learned  about the quadratic equation/formula?

    Answer me that, unanon yourself, then we will explain further. 

  • How do you think about the answers? You can sign in to vote the answer.
  • 1 month ago

    you have not listed any equations.

  • 1 month ago

    To be able to solve you need to equate then to zero 

    Hence 

    - 12d^2 + 2d - 30 = 0 

    - 4y^2 + 8y + 4 = 0 

    Top equ'n 

    Use quadratic eq'n 

    d = {- 2 +/- sqrt[(2^2 - 4(-12)(-30)]} / 2(-12)

    d = { - 2 +/- sqrt[ 4 - 1440]} / -24 

    d = { - 2 +/- sqrt[ - 1436] } / -24 

    This is unresolved , because you cannot take the square root of a negative number. , unless you move into IMGINARY numbers. 

    Bottom eq'n 

    Complete the Square 

    -4y^2 + 8y = -4 

    y^2 - 2y = 1 

    (y - 1^2 - ( - 1)^2 = 1 

    (y - 1)^2 -1 = 1 

    ( y - 1)^2 = 2 

    y - 1 = +/- sqrt(2)

    y = 1 +/- sqrt(2)

Still have questions? Get your answers by asking now.