Certainly! Let's solve the equation step by step:
Given equation:
[tex]\[ 5y^3 - 14y^2 - 3y = 0 \][/tex]
### Step 1: Factor out the Greatest Common Factor
First, notice that each term in the equation contains a factor of [tex]\( y \)[/tex]. We can factor [tex]\( y \)[/tex] out of the entire equation:
[tex]\[ y(5y^2 - 14y - 3) = 0 \][/tex]
### Step 2: Solve for the first solution
The equation now has two parts. One part is simply [tex]\( y = 0 \)[/tex].
[tex]\[ y = 0 \][/tex]
This is one of the solutions.
### Step 3: Solve the quadratic equation
We need to solve the quadratic equation found in the factored form:
[tex]\[ 5y^2 - 14y - 3 = 0 \][/tex]
To solve the quadratic equation, we will use the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 5 \)[/tex], [tex]\( b = -14 \)[/tex], and [tex]\( c = -3 \)[/tex].
### Step 4: Find the discriminant
Calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = (-14)^2 - 4(5)(-3) = 196 + 60 = 256 \][/tex]
### Step 5: Calculate the roots
Since the discriminant is positive, we have two distinct real roots:
[tex]\[ y = \frac{-(-14) \pm \sqrt{256}}{2 \cdot 5} = \frac{14 \pm 16}{10} \][/tex]
This results in two solutions:
[tex]\[ y = \frac{14 + 16}{10} = \frac{30}{10} = 3 \][/tex]
[tex]\[ y = \frac{14 - 16}{10} = \frac{-2}{10} = -\frac{1}{5} \][/tex]
### Summary
The complete set of solutions for the equation [tex]\( 5y^3 - 14y^2 - 3y = 0 \)[/tex] is:
[tex]\[ y = 0, 3, -\frac{1}{5} \][/tex]
