Answer
The solutions are $x = \frac{1}{2}$ and $x = 1$.
Work Step by Step
We have a quadratic equation, which is in the form $ax^2 + bx + c = 0$.
To factor the quadratic trinomial, look for the factors of the product $ac$ that when added together equals $b$.
The trinomial has $ac=2(1)=2$ and $b=-3$. The product is positive while the sum is negative. This means that both factors must be negative. Thus, the only possible factors are:
$-2$ and $-1$ (note that the sum is $-3$)
Rewrite the middle term of the trinomial using $-2$ and $-1$ to obtain:
$$2x^2 - 2x - x + 1 = 0$$
Group the first two terms together and group the last two terms together:
$$(2x^2 - 2x) + (-x - 1) = 0$$
Factor out the GCF in the first group and factor out $-1$ in the second group:
$$2x(x - 1) + (-1)(x - 1) = 0$$
Factor the GCF of the two groups, which is $x-1$:
$$(2x - 1)(x - 1) = 0$$
Solve the equation using the Zero-Product Property by equating each factor to $0$, then solve each equation.
First factor:
$2x - 1 = 0$
$2x = 1$
Divide each side by $2$:
$x = \frac{1}{2}$
Second factor:
$x - 1 = 0$
$x = 1$
The solutions are $x = \frac{1}{2}$ and $x = 1$.
