Question

A quadratic equation is shown below:

Answer 1

Part A:
3x^2-15x+20=0
ax^2+bx+c=0; a=3, b=-15, c=20
Radicand: R=b^2-4ac
R=(-15)^2-4(3)(20)
R=225-240
R=-15<0, then the equation has two imaginary solutions (no real)

Part B:
3x^2+5x-8=0
a=3, b=5, c=-8
R=(5)^2-4(3)(-8)
R=25+96
R=121>0, then the equation has two different real solutions:
x=[-b+-sqrt(R)] / (2a)
x=[-5+-sqrt(121)] / [2(3)]
x=(-5+-11) / 6

x1=(-5-11)/6=(-16)/6→x1=-8/3

x2=(-5+11)/6=6/6→x2=1

Solutions: x=-8/3 and x=1

I shose this method because I can get the solutions directly, and I don't have to guess the possible solutions