Without solving determine the number of real solutions for each quadratic equation b^2-4b+3=0

2n^2 + 7 = -4n + 5

x - 3x^2 = 5+ 2x - x^2
someone please help me!

b^2-4b+3=0

b²-3x-b+3=0

b(b-3)-1(b-3)=0

(b-3)(b-1)=0

either

b=3 or b=1

.

2n^2 + 7 = -4n + 5

2n²+4n+7-5=0

2n²+4n+2=0

2(n²+2n+1)=0

(n+1)²=0/2

:.n=-1

.

x - 3x^2 = 5+ 2x - x^2

0=5+ 2x - x^2-x +3x^2

0=5+x+2x²

2x²+x+5=0

comparing above equation with ax²+bx +c we get

a=2

b=1

c=5

x={-b±√(b²-4ac)}/2a ={-1±√(1²-4×2×5)}/2×1

={-1±√-39}/2

you just switch the numbers basically so it would be 3/17

6

step-by-step explanation: