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

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Answer 1 · 2016-02-24T14:37:35

$x=1/5,4$

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

Divide each term by $2$

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

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

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, $x=1/5,4$

How do you solve 10z^2+38z-8=010z^2+38z-8=0?