Question

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

Answer 1

`x=1/5,4`

Explanation:

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

Divide each term by `2`

`rarr5z^2+19z-4=0`

So, this is in the form of a Quadratic equation
(in form `ax^2+bx+c=0`)

Use Quadratic formula to solve

Quadratic formula:

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

Where,

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

Substitute the values

`rarrx=(-(19)+-sqrt(19^2-4(5)(-4)))/(2(5))`

`rarrx=(-19+-sqrt(361-4(-20)))/(10)`

`rarrx=(-19+-sqrt(361+80))/(10)`

`rarrx=(-19+-sqrt441)/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=1/5,4`