Question 35217
please help me find b^2 - 4ac and the number of real solutions to the equation below
-2x^2 + 5x - 6 = 0...........AX^2+BX+C=0..IS THE STD.EQN.COMPARING...
A=-2...B=5...........C=-6
B^2-4AC=5^2-4*(-2)*(-6)=25-48=-23=NEGATIVE...HENCE THERE ARE NO REAL ROOTS FOR THIS EQN.