# What are the x-intercepts of y = 2x^2 - 3x – 20

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### 4 Answers

y = 2x^2 - 3x -20

we need to determine x-intercept.

x-intercept is the point where the function intersect with the x-axis. In this case y value will be zero.

Then we need to calculate x values for the functin y to be zero.

y = 2x^2 - 3x - 20 = 0

Factor:

==> (2x +5)(x-4) = 0

==> x1= -2/5

==> x2= 4

Then y intersect with x-axis in two point:

(-2/5, 0) and (4, 0)

The x intercepts are, in fact, the roots of the equation:

2x^2 - 3x – 20 = 0 (when the graph of the y function is intercepting x axis).

Now, all we have to do is to solve the equation.

Because it's a quadratic, we'll apply the quadratic formula:

x1 = [-b+sqrt(b^2 - 4ac)]/2a

x2 = [-b-sqrt(b^2 - 4ac)]/2a

We'll identify a, b and c, which are the coefficients of the function:

y = ax^2 + bx + c

y = 2x^2 - 3x – 20

a = 2, b = -3, c = -20

Now, we'll calculate the roots:

x1 = [-(-3) + sqrt(9 +160)]/2

**x1 = (3+13)/2**

**x1 = 8**

**x2 = (3-13)/2**

**x2 = -5**

**So, the x intercepts are x1 and x2.**

y = 2x^2-3x-20.

To find the x intercepts, find values of x for which y = 0.

So 2x^2-3x-20 = 0.

We factorise 2x^2-3x-20 by splitting the middle term , -3x into -8x + 5x and grouping:

2x^2 - 8x + 5x - 20 = 0

2x(x-4) +5(x-4) = 0

(x-4)(2x+5) = 0

x-4 = 0 or 2x+5 = 0.

x-4 = 0 gives x = 4.

2x+5 = 0 gives x = -5/2

So the x intercepts of y = 2x^2-3x-20 = 0 are at x = 4 and x= -5/2 on the x axis.

To find the x-intercepts of y = 2x^2 - 3x - 20, we have to equate y = 0 and solve the resulting quadratic equation.

So if we substitute y=0 for the equation y = 2x^2 - 3x - 20 we get :

2x^2 - 3x - 20 = 0

=> 2x^2 - 8x + 5x - 20 = 0

=> 2x (x-4) + 5 (x-4) = 0

=> (2x+5) (x-4) = 0

=> x = -5/2 or x = 4

**Therefore the x - intercepts are at x = -5/2 and at x = 4 .**