How do you solve #3x^2+14x+15=0#?

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Answer 1 · 2018-04-20T13:55:47

#x = -5/3# or #x = -3#

#=> 3x^2 + 14x + 15 = 0#

It’s in the form of #ax^2 + bx + c = 0#

where,

Use formula for quadratic equation to find #x#

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

#x = (-14 +- sqrt(14^2 - (4 × 3 × 15)))/(2 × 3)#

#x = (-14 +- sqrt(196 - 180))/(6)#

#x = (-14 +- sqrt(16))/6#

#x = (-14 +-4)/6#

#x = (-14 + 4)/6 color(white)(....) "or" color(white)(....) x = (-14 - 4)/6#

#x = (-10)/6 color(white)(..........) "or" color(white)(....) x = (-18)/6#

#x = -5/3 color(white)(..........) "or" color(white)(....)x = -3#

Answer 2 · 2018-04-20T14:01:45

The quadratic formula is applicable on equations of the form:

#ax^2+bx+c=0#

And looks like this:

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

In our case:

#a=3#
#b=14#
#c=15#

#x= (-14+- sqrt((14)^2-4(3)(15)))/((2)(3))#

#x= (-14+- sqrt(196-180))/(6)#

#x= (-14+- sqrt(16))/(6)#

#x= (-14+- 4)/(6)#

So:

#x_1=(-18/6)=-3#

#x_2=(-10/6)=-5/3#