# How do you solve 3x^2 + 8x + 4 = 0 using the quadratic formula?

Oct 4, 2016

${x}_{\text{1,2}} = \left\{- \frac{2}{3} , - 2\right\}$

#### Explanation:

$\text{given : } 3 {x}^{2} + 8 x + 4 = 0$

$a {x}^{2} + b x + c = 0$

$a = 3$
$b = 8$
$c = 4$

$\Delta = \sqrt{{b}^{2} - 4 \cdot a \cdot c}$

$\Delta = \sqrt{{8}^{2} - 4 \cdot 3 \cdot 4}$

$\Delta = \sqrt{64 - 48}$

$\Delta = \sqrt{16}$

${x}_{\text{1"=(-b-Delta)/(2*a)" ; "x_1=(-8-4)/(2*3)" ; }} {x}_{1} = - \frac{12}{6} = - 2$

${x}_{2} = \frac{- b + \Delta}{2 \cdot a} \text{ ; "x_2=(-8+4)/(2*3)" ; } {x}_{2} = - \frac{4}{6} = - \frac{2}{3}$

${x}_{\text{1,2}} = \left\{- \frac{2}{3} , - 2\right\}$

Oct 4, 2016

color(green)(x=(-2,-2/3)

#### Explanation:

color(blue)(3x^2+8x+4=0

color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)

$\text{a,b and c are the coefficients of the terms}$

Assign the values

color(orange)(a=3

color(orange)(b=8

color(orange)(c=4

Solve

$\therefore x = \frac{- 8 \pm \sqrt{{8}^{2} - 4 \left(3\right) \left(4\right)}}{2 \left(3\right)}$

$\rightarrow \frac{- 8 \pm \sqrt{64 - 48}}{6}$

$\rightarrow \frac{- 8 \pm \sqrt{16}}{6}$

$\rightarrow \frac{- 8 \pm \sqrt{4 \cdot 4}}{6}$

$\rightarrow \frac{- 8 \pm 4}{6}$

We can divide the equation into two

color(violet)((-8-4)/6=-12/6=-2

color(purple)((-8+4)/6=-4/6=-2/3

$\therefore x = \left(- 2 , - \frac{2}{3}\right)$