Find the roots of each quadratic polynomial.

1. 16x + 40 + 2x^2

Solve. Round to the nearest hundredth.

2. 12 = 5x^2

1. A: No real solution.

2. A: ?

For number 2, can you show your work? What do you have so far?

2. 12 = 5x^2

12 / 5 = 5x / 5

12/5 = x^2

±√12/5 = x

That looks right to me. Are you allowed to use a calculator? What do you get for the value of x?

To find the roots of a quadratic polynomial, we need to set the polynomial equal to zero and use the quadratic formula. The quadratic formula states that for a quadratic equation in the form of ax^2 + bx + c = 0, the roots (or solutions) are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

Let's apply this formula to the given quadratic polynomials:

1. 16x + 40 + 2x^2
Rearranging the equation to the form of ax^2 + bx + c = 0, we have:
2x^2 + 16x + 40 = 0

Here, a = 2, b = 16, and c = 40.
Applying the quadratic formula, we have:

x = (-16 ± √(16^2 - 4(2)(40))) / (2(2))
x = (-16 ± √(256 - 320)) / 4
x = (-16 ± √(-64)) / 4

Since the discriminant (represented by b^2 - 4ac) is negative, the quadratic equation does not have any real roots. In this case, we can say that the roots are complex numbers.

2. 12 = 5x^2
Rearranging the equation to the form of ax^2 + bx + c = 0, we have:
5x^2 - 12 = 0

Here, a = 5, b = 0, and c = -12.
Applying the quadratic formula, we have:

x = (0 ± √(0^2 - 4(5)(-12))) / (2(5))
x = (0 ± √(0 + 240)) / 10
x = (0 ± √240) / 10

Now, we need to simplify the square root of 240:
√240 = √(16 * 15) = √16 * √15 = 4√15

Therefore, the roots of the quadratic polynomial 12 = 5x^2 are:
x = ±(4√15) / 10
Simplifying further:
x = ±2√15 / 5

So, the roots are ±2√15 / 5.