# How do you solve 10z^2+38z-8=0?

Feb 24, 2016

$x = \frac{1}{5} , 4$

#### Explanation:

color(blue)(10z^2+38z-8=0

Divide each term by $2$

$\rightarrow 5 {z}^{2} + 19 z - 4 = 0$

So, this is in the form of a Quadratic equation
(in form $a {x}^{2} + b x + c = 0$)

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

Where,

In this case, color(red)(a=5,b=19,c=-4

Substitute the values

$\rightarrow x = \frac{- \left(19\right) \pm \sqrt{{19}^{2} - 4 \left(5\right) \left(- 4\right)}}{2 \left(5\right)}$

$\rightarrow x = \frac{- 19 \pm \sqrt{361 - 4 \left(- 20\right)}}{10}$

$\rightarrow x = \frac{- 19 \pm \sqrt{361 + 80}}{10}$

$\rightarrow x = \frac{- 19 \pm \sqrt{441}}{10}$

rarrcolor(orange)(x=(-19+21)/(10)

Now we have two values for $x$

1)color(pink)(x=(-19+21)/10

2)color(pink)(x=(-19-21)/10

Solve for the two values:

rArr1)color(green)(x=(-19+21)/10=2/10=1/5

rArr2)color(green)(x=(-19-21)/10=-40/10=-4

So, $x = \frac{1}{5} , 4$