Solve 5x² - 6x - 3 ≤ |8x|

please show how to do it as well. thanks

We have to consider two possible cases. This is generally the case when you have to solve equations or inequalities involving absolute values.
* Case 1: 8x ≥ 0:
This means your inequality becomes:
5x^2 - 14x - 3 ≤ 0
Solving the quadratic inequality gives: -1/5 ≤ x ≤ 3. Given that 8x ≥ 0, this means that x must be in [0,3].
* Case 2: 8x < 0:
Now we get:
5x^2 + 2x - 3 ≤ 0.
Solving the quadratic inequality gives:
-1 ≤ x ≤ 3/5. Given that 8x < 0, this means x must be in [-1,0).
Combining the two scenarios gives us the solution: x is in the range [-1, 3]. A graphical plot of the function confirms this.

Dear Kathy,

5x^(2)-6x-3<=|8x|

Since |8x| is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.
|8x|>=5x^(2)-6x-3

Remove the absolute value term. This creates a \ on the right-hand side of the equation because |x|=\x.
8x>=\(5x^(2)-6x-3)

Set up the + portion of the \ solution.
8x>=5x^(2)-6x-3

Solve the first equation for x.
-(1)/(5)<=x<=3

Set up the - portion of the \ solution. When solving the - portion of an inequality, flip the direction of the inequality sign.
8x<=-(5x^(2)-6x-3)

Since x is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.
-(5x^(2)-6x-3)>=8x

Multiply -1 by each term inside the parentheses.
-5x^(2)+6x+3>=8x

Move all terms not containing x to the right-hand side of the inequality.
-5x^(2)-2x+3>=0

Multiply each term in the equation by -1.
5x^(2)+2x-3<=0

Remove the fraction by multiplying the first term of the factor by the denominator of the second term.
(x+1)(5x-3)<=0

Set each of the factors of the left-hand side of the inequality equal to 0 to find the critical points.
x+1=0_5x-3=0