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How many solutions does this have y=4x^2+20x-5

User Lambinator
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Answer:

The equation y = 4x^2 + 20x - 5 represents a quadratic equation. To determine the number of solutions it has, we can use the discriminant, which is the part of the quadratic formula under the square root symbol (√). The discriminant is calculated as follows:

Discriminant (D) = b^2 - 4ac

In the equation y = 4x^2 + 20x - 5, we have a = 4, b = 20, and c = -5. Plugging these values into the discriminant formula, we get:

D = (20)^2 - 4(4)(-5)

D = 400 + 80

D = 480

The discriminant value is 480.

Now, let's see what the discriminant tells us about the solutions of the equation:

1. If the discriminant is positive (D > 0), then the equation has two distinct real solutions.

2. If the discriminant is zero (D = 0), then the equation has one real solution (also called a double root).

3. If the discriminant is negative (D < 0), then the equation has no real solutions, but it may have complex solutions.

In our case, the discriminant is positive (D = 480), so the equation has two distinct real solutions.

To find the solutions, we can use the quadratic formula:

x = (-b ± √D) / (2a)

Plugging in the values, we get:

x = (-20 ± √480) / (2 * 4)

x = (-20 ± √480) / 8

x = (-20 ± √(16 * 30)) / 8

x = (-20 ± 4√30) / 8

x = (-5 ± √30) / 2

So, the equation y = 4x^2 + 20x - 5 has two distinct real solutions given by x = (-5 + √30) / 2 and x = (-5 - √30) / 2.

User Martinus
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