# How do you solve 2x^2+5x+2=0?

Nov 26, 2015

The solutions are:

 color(blue)(x=-1/2 color(white)(...........)

 color(blue)(x=-2

#### Explanation:

$2 {x}^{2} + 5 x + 2 = 0$

We can Split the Middle Term of this expression to factorise it and then find the solutions.

In this technique, if we have to factorise an expression like $a {x}^{2} + b x + c$, we need to think of 2 numbers such that:

${N}_{1} \cdot {N}_{2} = a \cdot c = 2 \cdot 2 = 4$
AND
${N}_{1} + {N}_{2} = b = 5$

After trying out a few numbers we get ${N}_{1} = 4$ and ${N}_{2} = 1$
$4 \cdot 1 = 4$, and $4 + 1 = 5$

$2 {x}^{2} + \textcolor{b l u e}{5 x} + 2 = 2 {x}^{2} + \textcolor{b l u e}{4 x + 1 x} + 2 = 0$

$2 x \left(x + 2\right) + 1 \left(x + 2\right) = 0$

color(blue)((2x+1)(x+2)=0

Now we equate the factors to zero:

2x+1=0, color(blue)(x=-1/2

x+2=0, color(blue)(x=-2