Question

Solve the equation : 21x2 -28x +10=0.

Answer 1

21x^(2)-28x+10=0

x=(-b+- sqrt(b^(2)-4ac))/(2a) where ax^(2)+bx+c=0

x=(-(-28)\ sqrt((-28)^(2)-4(21)(10)))/(2(21))

x=(28+- sqrt((-28)^(2)-4(21)(10)))/(2(21))

x=(28+-2i sqrt(14))/2(21)

x=(28+-2i sqrt(14))/(42)
---------------------------------------
x=(14+i sqrt(14))/21

x=(14-i sqrt(14))/21
---------------------------------------
x=(14+i sqrt(14))/(21) , (14-i sqrt(14))/(21)

Answer 2

straight for the juggular ....