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

- 12d^2 + 2d - 30

- 4y^2 + 8y + 4

### 6 Answers

- PhilomelLv 71 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)

- PinkgreenLv 71 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.

- DavidLv 71 month ago
Without an equality sign that equals something the given expressions are not equations.

- rotchmLv 71 month ago
Hint: Ever heard/learned about the quadratic equation/formula?

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

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- lenpol7Lv 71 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)