Question

find all solutions by factoring.
2c² - 9c = 5

Explanation:

Step1: Rewrite in standard quadratic form

First, we need to rewrite the equation \(2c^{2}-9c = 5\) in the standard quadratic form \(ax^{2}+bx + c=0\). Subtract 5 from both sides:
\(2c^{2}-9c - 5=0\)

Step2: Factor the quadratic

We need to factor the quadratic expression \(2c^{2}-9c - 5\). We look for two numbers that multiply to \(2\times(- 5)=-10\) and add up to \(-9\). The numbers are \(-10\) and \(1\).
We rewrite the middle term using these numbers:
\(2c^{2}-10c + c-5 = 0\)
Now, factor by grouping:
\(2c(c - 5)+1(c - 5)=0\)
Then, we can factor out \((c - 5)\):
\((2c + 1)(c - 5)=0\)

Step3: Solve for \(c\)

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
Case 1: \(2c+1 = 0\)
Subtract 1 from both sides: \(2c=-1\)
Divide both sides by 2: \(c=-\frac{1}{2}\)
Case 2: \(c - 5=0\)
Add 5 to both sides: \(c = 5\)

Answer:

The solutions are \(c =-\frac{1}{2}\) and \(c = 5\)