Answer
The solutions are $x = \frac{1}{2}, - \frac{5}{2}$.
Work Step by Step
Rewrite the equation so that all terms are on the left side of the equation:
$$4x^2 + 8x - 5 = 0$$
To factor a quadratic trinomial in the form $ax^2 + bx + c$, we look at factors of the product of $a$ and $c$ that when added together equals $b$.
For the trinomial $4x^2 + 8x - 5$, $(a)(c)$ is $(4)(-5)$, or $-20$, so we have to look for factors of $-20$ that when added together will equal $b$ or $8$. One of the factors needs to be positive and the other one negative, but the positive factor should be the one with the greater absolute value. This is because a negative number multiplied with a positive number will equal a negative number; however, when a negative number is added to a positive number, the result can be either negative or positive, depending on which number has the greater absolute value.
The following possibilities are:
$-20(20)(-1)$
$20+(-1) = 19$
$-20=(10)(-2)$
$10+(-2) = 8$
$-20=(5)(-4)$
$5+(-4) = 1$
The second pair, $10$ and $-2$, is the one we are looking for.
Use this factors to split the middle term of the trinomial:
$$4x^2 + 10x - 2x - 5$$
Factor by grouping. Group the first two terms together and the last two terms together:
$$(4x^2 + 10x) - (2x + 5) = 0$$
$2x$ is a common factor for the first group, so factor it out:
$$2x(2x + 5) - (2x + 5)$$
$2x + 5$ is a common of both groups, so factor it out.:
$$(2x - 1)(2x + 5)$$
Thus, the equation above is equivalent to:
$$(2x - 1)(2x + 5)=0$$
Use the Zero-Product Property by equating each factor to zero, then solve each equation.
For the first factor:
$2x - 1 = 0$
$2x = 1$
Divide each side by $2$ to solve for $x$:
$x = \frac{1}{2}$
For the second factor:
$2x + 5 = 0$
$2x = -5$
Divide each side by $2$ to solve for $x$:
$x = -\frac{5}{2}$
The solutions are $x = \frac{1}{2}, - \frac{5}{2}$.
To check if the solutions are correct, substitute each on into the equation.
When $x = \frac{1}{2}$:
$4x^2=-8x+5\\
4\left(\frac{1}{2}\right)^2 =-8\left(\frac{1}{2}\right) +5\\
4\left(\frac{1}{4}\right)=-4+5\\
1=1$
When $x = -\frac{5}{2}$:
$4x^2=-8x+5\\
4\left(- \frac{5}{2})\right)^2=-8\left(-\frac{5}{2}\right) +5\\
4\left(\frac{25}{4}\right)=20+5\\
25=25$
Therefore, the solutions are $\frac{1}{2}$ and $-\frac{5}{2}$.
