Question

PLEASE HELP ME NOW URGENT

Answer 1

ANSWER

$y = 3 \pm \sqrt{21}$

EXPLANATION

The quadratic equation is:

${y}^{2} - 6y - 12 = 0$

Group variable terms:

${y}^{2} - 6y = 12$

Add the square of half, the coefficient of y to both sides.

${y}^{2} - 6y + ( - 3) ^{2} = 12 + ( - 3) ^{2}$

${y}^{2} - 6y + 9= 12 + 9$

The LHS us now a perfect square trinomial:

${(y - 3)}^{2}= 21$

Take square root:

$y - 3 = \pm \sqrt{21}$

$y = 3 \pm \sqrt{21}$

The first choice is correct.

Answer 2

3±√21. The equation $y^{2}-6y-12=0$ has two possible solutions 3+√21 y 3-√21.

If we have a general quadratic equation $ay^{2} +by+c=0$ we can solves the equation by completing the square. First, we divide the quadratic equation by a, we obtain $y^{2} +\frac{b}{a} y+\frac{c}{a} =0$.

For this problem, we have $y^{2}-6y-12=0$

We can skipped division in this example since the coefficient of $x^{2}$ is 1.

Move the term c to the right side of the equation

$y^{2}-6y=12$

Completing the square on the left side of the equation and balance this by adding the same number to the right side of the equation, with b = -6.

$(\frac{b}{2})^{2} =(\frac{-6}{2})^{2}=(-3)^{2} =9$

$y^{2}-6y+9=12+9$

$(y-3)^{2}=21$

Take the square root on both sides of the equation:

y - 3 = ±√21

Add 3 from both sides:

y = 3 ± √21