How do you solve #t^2 + 5t - 24 = 0 #?

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Answer 1 · 2016-03-26T06:43:15

#x=3,-8#

#color(blue)(t^2+5t-24=0#

This is a Quadratic equation (in form #ax^2+bx+c=0#)

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

Where

#color(red)(a=1,b=5,c=-24#

#rarrx=(-5+-sqrt(5^2-4(1)(-24)))/(2(1))#

#rarrx=(-5+-sqrt(25-(-96)))/(2)#

#rarrx=(-5+-sqrt(25+96))/(2)#

#rarrx=(-5+-sqrt(121))/(2)#

#rarrx=(-5+-11)/(2)#

Now we have #2# solutions

#color(purple)(x=(-5+11)/(2)=6/2=3#

#color(indigo)(x=(-5-11)/(2)=-16/2=-8#

#:. color(blue)( ul bar |x=3,-8|#