# How many solutions does 4x^2 + 8x + 3 = 0 have?

May 1, 2015

The easiest way to determine the answer to your question is to evaluate the $\textcolor{red}{\text{discriminant}}$
For a quadratic in the form:
$y = a {x}^{2} + b x + c$
the color(red)("discriminant") is equal to
$\textcolor{red}{{b}^{2} - 4 a c}$

$4 {x}^{2} + 8 x + 3 = 0$
The $\textcolor{red}{\text{discriminant}}$ is
$\Delta = \frac{{8}^{2} - 4 \left(4\right) \left(3\right)}{2 \left(4\right)}$

$\Delta = 2$

The number of solutions is determined by the value of the $\textcolor{red}{\text{discriminant, } \Delta}$

$\Delta \left\{\begin{matrix}> 0 \text{ two solutions" \\ =0" one solution" \\ <0" no solutions}\end{matrix}\right.$

So
$4 {x}^{2} + 8 x + 3 = 0$ has two solutions

Further explanation:
The discriminant comes from the formula for quadratic root solutions
$\frac{- b \pm \sqrt{\textcolor{red}{{b}^{2} - 4 a c}}}{2 a}$
and it should be fairly obvious from this to understand why

$\textcolor{red}{{b}^{2} - 4 a c} \left\{\begin{matrix}> 0 \rightarrow \text{ two solutions" \\ =0 rarr" one solution" \\ <0 rarr " no solutions}\end{matrix}\right.$