# How do you solve the quadratic with complex numbers given 3t^2+8t+5=-2t^2?

Mar 17, 2018

$t = - \frac{4}{5} + \frac{3}{5} i \textcolor{w h i t e}{. .} \mathmr{and} \textcolor{w h i t e}{. .} t = - \frac{4}{5} - \frac{3}{5} i$

#### Explanation:

Add $2 {t}^{2}$ to both sides

$5 {t}^{2} + 8 t + 5 = 0$

You can either complete the square or use the 'formula'

Keeping the standard form I will use $x$

$y = a {x}^{2} + b x + c \implies x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

When I first came across these I made a point of writing the above at the start of every solution. It has stuck in my memory ever since. That was a long time ago so the method works!

a=+5;b=+8 and c=+5 giving, in this case $t$:

$t = \frac{- 8 \pm \sqrt{{8}^{2} - 4 \left(5\right) \left(5\right)}}{2 \left(5\right)}$

$t = - \frac{8}{10} \pm \frac{\sqrt{- 36}}{10}$

Bit $- 36$ is the same as $36 \times \left(- 1\right)$

$t = - \frac{8}{10} \pm \left[\frac{\sqrt{36} \textcolor{w h i t e}{.} \sqrt{- 1}}{10}\right]$

But $\sqrt{36} = 6 \mathmr{and} \sqrt{- 1} = i$

$t = - \frac{4}{5} \pm \frac{3}{5} \textcolor{w h i t e}{.} i$

$t = - \frac{4}{5} + \frac{3}{5} i \textcolor{w h i t e}{. .} \mathmr{and} \textcolor{w h i t e}{. .} t = - \frac{4}{5} - \frac{3}{5} i$

Mar 17, 2018

${t}_{1} = - \frac{4}{5} + i \frac{3}{5} \mathmr{and} {t}_{2} = - \frac{4}{5} - i \frac{3}{5}$

#### Explanation:

$3 {t}^{2} + 8 t + 5 = - 2 {t}^{2} | + 2 {t}^{2}$
5t^2+8t+5=0|:5
${t}^{2} + \frac{8}{5} t + 1 = 0$

${\left(t + \frac{4}{5}\right)}^{2} + 1 - \frac{16}{25} = 0 | - \frac{9}{25} | \sqrt{}$
$t + \frac{4}{5} = \pm \sqrt{- \frac{9}{25}} | - \frac{4}{5}$
$t = - \frac{4}{5} \pm \sqrt{- \frac{9}{25}}$

Determinating the complex root...

$\sqrt{- a} = i \cdot \sqrt{a}$

$\pm \sqrt{- \frac{9}{25}} = \pm i \frac{3}{5}$
${t}_{1} = - \frac{4}{5} + i \frac{3}{5} \mathmr{and} {t}_{2} = - \frac{4}{5} - i \frac{3}{5}$