Solve the quadratic equation?

0 ⤊

Solve the quadratic equation?

7z^(2)+17z+21=0

2006-11-22 06:30:11 · 4 answers · asked by oss o 1 in Science & Mathematics ➔ Engineering

4 answers

z=-17/14+1/14*i*299^(1/2), -17/14-1/14*i*299^(1/2)

This is because using the quadratic formulae gives sqrt(-299)
and to solve this one needs to know about i where i=sqrt(-1)

i is a complex number thus the answers for z are complex numbers

2006-11-22 06:35:10 · answer #1 · answered by Oz 4 · 0⤊ 0⤋

14z+17z+21=0
31z+21=0

2006-11-22 14:39:28 · answer #2 · answered by Robby 1 · 0⤊ 2⤋

root 1 = (-17+((289-4*7*21)^0.5)/14 = -17/14 +i 17.2/14

root 2 = (-17*((289-4*7*21)^0.5)/14 = -17/14 -i 17.2/14

2006-11-22 14:39:56 · answer #3 · answered by maussy 7 · 0⤊ 1⤋

z = -1.2142 + 1.2355i and z = -1.2142 - 1.2355i

2006-11-22 14:47:42 · answer #4 · answered by David R 1 · 0⤊ 0⤋