How do you use the quadratic formula to solve #3x ^ { 2} + 3x - 5= 0#?

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Answer 1 · 2018-04-11T13:32:41

#=> x = { ( -3 + sqrt(69) ) / (6) , ( -3 - sqrt(69) ) / (6) } #

Or approximately

#=> x approx { 0.884 , -1.884 } #

The quadratic is #ax^2 + bx +c = 0 #

and the formula is: #x =( -b pm sqrt(b^2 - 4ac ) ) / (2a ) #

In this case #a = 3 # , # b = 3 # and #c = -5 #

#=> x = ( -3 pm sqrt( 3^2 - (4*3*(-5) ) ) ) / (2*3)#

#=> x = ( -3 pm sqrt(69) ) / (6) #

#=> x = { ( -3 + sqrt(69) ) / (6) , ( -3 - sqrt(69) ) / (6) } #

Or approximately

#=> x approx { 0.884 , -1.884 } #

Answer 2 · 2018-04-11T13:44:40

#x=(−3+sqrt69)/(6)=0.88#
or
#x=(−3-sqrt69)/(6)=-1.88#

The equation #3x^2=3x-5=0# is written in the form #y=ax^2+bx+c#, so #a=3, b=3, c=-5#

The quadratic formula is #x=(−b±sqrt(b^2−4ac))/(2a)#

Substitute the values of a, b and c into the formula

#x=(−(3)±sqrt(3^2−4(3xx-5)))/(2(3))#
#x=(−3±sqrt(9+60))/(6)#
#x=(−3±13)/(6)#

#x=(−3+sqrt69)/(6)=0.88#
or
#x=(−3-sqrt69)/(6)=-1.88#