Alvin throws the football to a receiver who jumps up to catch the ball. The height of the ball over time can be represented by the quadratic equation -4.9t^2 + 7.5t + 1.8 = 2.1 . This equation is based on the acceleration of gravity -4.9 m/s^2, the velocity of his pass is 7.5 m/s and releases the football at a height of 1.8 meters, and the height where the receiver catches the ball of 2.1 meters. Put the equation in standard form and then solve by using the quadratic equation.

I don't understand any of this, could someone help me figure out how to put it in both the standard form and quadratic equation.

-4.9t^2 + 7.5t + 1.8 = 2.1

we usually start with a positive square term

4.9t^2 - 7.5t + 0.3 = 0

for a quadratic in the standard form of
ax^2 + bx + c = 0
x = (-b ± √(b^2 - 4ac) )/(2a)

sub in the corresponding values from my equation and find x

Okay, I did what you said and for x I got= -5.43152 now do I sub that into the top equation but instead of x I put it in for t?

To put the quadratic equation in standard form, we need to rearrange the equation so that all terms are on one side and set equal to zero. Let's start by subtracting 2.1 from both sides of the equation:

-4.9t^2 + 7.5t + 1.8 - 2.1 = 0

Simplifying, we get:

-4.9t^2 + 7.5t - 0.3 = 0

Now that the equation is in standard form, we can use the quadratic formula to solve for t. The quadratic formula is:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = -4.9, b = 7.5, and c = -0.3. Plugging these values into the formula:

t = (-7.5 ± √(7.5^2 - 4(-4.9)(-0.3))) / (2(-4.9))

Simplifying further:

t = (-7.5 ± √(56.25 - 5.88)) / (-9.8)

t = (-7.5 ± √50.37) / (-9.8)

Now, we can calculate the two possible values of t:

t1 = (-7.5 + √50.37) / (-9.8)
t2 = (-7.5 - √50.37) / (-9.8)

Evaluating these expressions using a calculator, we get:

t1 ≈ 0.776 seconds
t2 ≈ -0.073 seconds

Since time cannot be negative in this context, we discard the negative value t2. Therefore, the time it takes for the receiver to catch the ball is approximately 0.776 seconds.

Sure! Let's break down the steps to put the equation in standard form and then solve using the quadratic equation.

Step 1: Convert the equation to standard form.
The standard form of a quadratic equation is written as: ax^2 + bx + c = 0.
To convert the given equation to standard form, we need to move all the terms to one side so that the equation equals zero.

Starting with -4.9t^2 + 7.5t + 1.8 = 2.1, subtract 2.1 from both sides:
-4.9t^2 + 7.5t + 1.8 - 2.1 = 0
-4.9t^2 + 7.5t - 0.3 = 0

Now, the equation is in standard form: -4.9t^2 + 7.5t - 0.3 = 0.

Step 2: Solve the equation using the quadratic formula.
The quadratic formula is x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients from the equation in standard form.

In our case, a = -4.9, b = 7.5, and c = -0.3.

Plug these values into the quadratic formula:
t = (-7.5 ± √(7.5^2 - 4(-4.9)(-0.3))) / (2(-4.9))
t = (-7.5 ± √(56.25 - 5.88)) / (-9.8)
t = (-7.5 ± √50.37) / (-9.8)

Now, we have two possible solutions for t: t = (-7.5 + √50.37) / (-9.8) (call it t1) and t = (-7.5 - √50.37) / (-9.8) (call it t2).

Finally, you can use a calculator to evaluate the square root and perform the necessary calculations to find the values of t1 and t2.