R Ryan Aug 23, 2004 #1 Does anyone know what would be the most efficient way to solve a quadratic equation, by either using solver or not?
Does anyone know what would be the most efficient way to solve a quadratic equation, by either using solver or not?
V Vasant Nanavati Aug 23, 2004 #2 =(-B1+((B1^2)-(4*A1*C1)))/(2*A1) and =(-B1-((B1^2)-(4*A1*C1)))/(2*A1) where A1, B1 and C1 contain the respective coefficients.
=(-B1+((B1^2)-(4*A1*C1)))/(2*A1) and =(-B1-((B1^2)-(4*A1*C1)))/(2*A1) where A1, B1 and C1 contain the respective coefficients.
P Phil Aug 23, 2004 #3 Looks like you forgot the sqrt ! (-B + sqrt(B^2-4AC)) /2A & (-B - sqrt(B^2-4AC)) /2A
H hgrove Aug 23, 2004 #4 Ryan wrote... Does anyone know what would be the most efficient way to solve a quadratic equation, by either using solver or not? Click to expand... See http://www.google.com/groups?selm=281020022338554689%[email protected]
Ryan wrote... Does anyone know what would be the most efficient way to solve a quadratic equation, by either using solver or not? Click to expand... See http://www.google.com/groups?selm=281020022338554689%[email protected]
V Vasant Nanavati Aug 24, 2004 #5 I most certainly did! Thanks. -- Vasant Phil said: Looks like you forgot the sqrt ! (-B + sqrt(B^2-4AC)) /2A & (-B - sqrt(B^2-4AC)) /2A Click to expand...
I most certainly did! Thanks. -- Vasant Phil said: Looks like you forgot the sqrt ! (-B + sqrt(B^2-4AC)) /2A & (-B - sqrt(B^2-4AC)) /2A Click to expand...
