How do you solve #-\frac { 1} { 2} x ^ { 2} + 3x - 3= 0#?

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Answer 1 · 2017-03-18T20:23:43

#"answer : "x_1=sqrt3+3" , "x_1=-sqrt3+3#

#-1/2 x^2+3x-3=0#

#"We can calculate using the formula below."#

#ax^2+bx+c=0 "(general formula for quadratic equations.)"#

#Delta=sqrt (b^2-4*a*c)#

#"Where "a=-1/2 " , "b=3 " , "c=-3 #

#Delta=sqrt(3^2-4*1/2*3)=sqrt(9-6)=sqrt 3#

#x_1=(-b-Delta)/(2*a)=(-3-sqrt3)/(cancel(2)*(-1/cancel(2)))#

#x_1=sqrt3+3#

#x_2=(-b+Delta)/(2*a)=(-3+sqrt3)/(cancel(2)*(-1/cancel(2)))#

#x_1=-sqrt3+3#