Using the quadratic formula to solve equations.

-12^2 + 12 x-9 = -7x^2
10x^2 + 5x + 12 = 0
6x^2 + 12x = - 1
11x^2 - 7x -1 = 0
2x^2 -18x + 16 = 0

Can u show work to help me out

I will do one .

11x^2-7x-1=0
a= 11
b=-7
c=-1

x= (-b+-sqrt(b^2-4ac)/2a

Put the numbers in, and solve.

= 7 +or- isqroot 5/22

Certainly! I can help you solve these quadratic equations using the quadratic formula, which is given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

Let's go through each equation step by step:

1. -12^2 + 12x - 9 = -7x^2

First, rearrange the equation to bring all terms to one side:

-7x^2 - 12x + 144 - 9 = 0

Rearrange again to match the quadratic equation form (ax^2 + bx + c = 0):

-7x^2 - 12x + 135 = 0

Now we can identify the coefficients in the quadratic formula:
a = -7, b = -12, c = 135

Substituting these values into the quadratic formula:

x = (-(-12) ± √((-12)^2 - 4(-7)(135))) / (2(-7))

Simplifying:

x = (12 ± √(144 + 3780)) / (-14)

x = (12 ± √3924) / (-14)

x = (12 ± √(4 * 981)) / (-14)

x = (12 ± 2√981) / (-14)

x = (6 ± √981) / (-7)

The solutions are x = (6 + √981) / (-7) and x = (6 - √981) / (-7).

2. 10x^2 + 5x + 12 = 0

In this equation, we already have it in the quadratic equation form:

a = 10, b = 5, c = 12

Substituting these values into the quadratic formula:

x = (-(5) ± √((5)^2 - 4(10)(12))) / (2(10))

Simplifying:

x = (-5 ± √(25 - 480)) / 20

x = (-5 ± √(-455)) / 20

Since the square root of a negative number is not a real number, this equation has no real solutions.

3. 6x^2 + 12x = -1

Rearranging the equation:

6x^2 + 12x + 1 = 0

Now we can identify the coefficients in the quadratic formula:
a = 6, b = 12, c = 1

Substituting these values into the quadratic formula:

x = (-(12) ± √((12)^2 - 4(6)(1))) / (2(6))

Simplifying:

x = (-12 ± √(144 - 24)) / 12

x = (-12 ± √120) / 12

x = (-12 ± √(4 * 30)) / 12

x = (-12 ± 2√30) / 12

x = (-6 ± √30) / 6

The solutions are x = (-6 + √30) / 6 and x = (-6 - √30) / 6.

4. 11x^2 - 7x - 1 = 0

Again, we have it in the quadratic equation form:

a = 11, b = -7, c = -1

Substituting these values into the quadratic formula:

x = (-(7) ± √((-7)^2 - 4(11)(-1))) / (2(11))

Simplifying:

x = (-7 ± √(49 + 44)) / 22

x = (-7 ± √93) / 22

The solutions are x = (-7 + √93) / 22 and x = (-7 - √93) / 22.

5. 2x^2 - 18x + 16 = 0

Again, arranging the equation:

2x^2 - 18x + 16 = 0

Identifying the coefficients:
a = 2, b = -18, c = 16

Substituting into the quadratic formula:

x = (-(18) ± √((-18)^2 - 4(2)(16))) / (2(2))

Simplifying:

x = (-18 ± √(324 - 128)) / 4

x = (-18 ± √196 ) / 4

x = (-18 ± 14) / 4

The solutions are x = (-18 + 14) / 4 and x = (-18 - 14) / 4.

x = -1 and x = -8.

I hope this helps! If you have any further questions, feel free to ask.