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#