Solve the inequality 2x^2<12-5x

2x^2 + 5x -12 < 0

(2x-3)(x+4) < 0

Take it from here.

x> 2

To solve the inequality 2x^2 < 12 - 5x, we can follow these steps:

Step 1: Move all terms to one side of the inequality to create a quadratic equation in standard form.
2x^2 + 5x - 12 < 0

Step 2: Find the roots (zeros) of the quadratic equation by factoring, using the quadratic formula, or completing the square.
Since factoring may not always work, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 2, b = 5, and c = -12.
x = (-5 ± √(5^2 - 4(2)(-12))) / (2(2))
Simplifying further, we have:
x = (-5 ± √(25 + 96)) / (4)
x = (-5 ± √121) / 4
x = (-5 ± 11) / 4

So, we have two possible values of x:
x = (-5 + 11) / 4 = 6/4 = 3/2
x = (-5 - 11) / 4 = -16/4 = -4

Step 3: Plot the roots on a number line to divide the line into regions.
We have two roots, x = 3/2 and x = -4. So, let's plot them on a number line:

-∞ -4 3/2 +∞
-----------------|--------|-------|----------------------

Step 4: Choose a test point in each region and substitute it into the inequality to determine the sign of the expression.
For simplicity, we can choose 0 as a test point in the regions:
Region 1: x < -4
Substituting x = 0, we have 2(0)^2 + 5(0) - 12 < 0, which simplifies to -12 < 0.
Since -12 is indeed less than 0, the inequality is true in this region.

Region 2: -4 < x < 3/2
Substituting x = 0, we have 2(0)^2 + 5(0) - 12 < 0, which again simplifies to -12 < 0.
So, the inequality is true in this region as well.

Region 3: x > 3/2
Substituting x = 0, we have 2(0)^2 + 5(0) - 12 < 0, resulting in -12 < 0.
However, in this region, x > 3/2, which means any value larger than 3/2 would make the quadratic expression positive. Therefore, the inequality is false in this region.

Step 5: Write the final solution.
The solution to the inequality 2x^2 < 12 - 5x is:
x < -4 or -4 < x < 3/2