1.)Complete the square to find all real solutions of equation 3 x^2 + 4 x - 1 = 0.
X1=
X2=
2.)\sqrt{7 x - 9} + 6 = 3x
there are two solutions
3.)4 <= 5 x - 4 <10
4.)(x/4x-16)-7 = (1/x-4)
5.) (1/R) = (1/R1)+ (1/R2)
for R1
I'll do the first one. Will be glad to critique if you show the work for the rest.
3 x^2 + 4 x - 1 = 0
divide by 3 throughout:
x² + (4/3)x - (1/3) = 0
(x+2/3)²-(2/3)²-(1/3)=0
(x+2/3)² = 7/9
x+2/3 = ±(√7)/3
x = -2/3 ±(√7)/3
Note: you'd be better off posting the questions in separate posts. There will be more chance that some of them get answered sooner.