Algebra 2a HELP ME? Plz help me asap I need this :(

x²-8x+2 =-10

3x²+6x-15=30


3x²+12x+20=11

2x²+12x=80

4x²-56+80=-122

-2x²-x+10=-18




Help me plz I don't get it:(
Need to know asap plz:( I can use ur help:(

depends how you want to solve it.... first couple i'm going to solve by completing the square


x^2 -8x = -10 - 2

1(x^2 -8x) = -12

1(x^2 -8x + 4^2 -4^2) = -12

1(x^2 -8x +16 - 16) = -12

1(x^2 -8x +16) + 1(-16) = -12

1(x-4)(x-4) - 16 = -12

1(x-4)^2 = -12 + 16

(x-4)^2 = 4

√(x-4)^2 = +/- √4

x - 4 = +/- 2

x = 4 +/- 2

x1 = 4 - 2

x1 = 2

x2 = 4 + 2

x2 = 6

THEREFORE x= 2, 6






3x^2 +6x = 30 + 15

3(x^2 +2x) = 45

3(x^2 +2x +1^2 - 1^2) = 45

3(x^2 +2x +1 -1) = 45

3(x^2 +2x +1) + 3(-1) = 45

3(x+1)(x+1) - 3 = 45

3(x+1)^2 = 45 + 3

3(x+1)^2 = 48

3(x+1)^2 / 3 = 48/3

(x+1)^2 = 16

√(x+1)^2 = +/- √16

x + 1 = +/- 4

x = -1 +/- 4

x1 = -1 - 4

x1 = -5

x2 = -1 + 4

x2 = 3

THEREFORE x = -5, 3








solving the next one by factoring

3x^2 +12x + 20 - 11 = 0

3x^2 +12x + 9 = 0

3(x^2 +4x +3) = 0

3(x+1)(x+3) = 0

(x+1)=0 and (x+3)=0

x+1 =0 and x+3 = 0

x = -1 and x = -3

THEREFORE x = -3, -1





same with the next one.

2x^2 +12x - 80 = 0

2(x^2 +6x - 40) = 0

2(x+10)(x-4) = 0

(x-4)=0 and (x+10) = 0

x= 4 and x = -10

THEREFORE x = -10, 4






Solving the last 2 by quadratic formula.

4x^2 -56x + 80 + 122 = 0

4x^2 -56x + 202 = 0 <-----a = 4, b = -56, c= 202

x = (-b +/- √(b^2-4ac)) /2a

x = (-(-56) +/- √((-56)^2 -4(4)(202))) / 2(4)

x = (56 +/- √(3136 -(3232))) / 8

x = (56 +/- √-96) / 8

x = (56 +/- (√16 times √-1 times √6)) / 8

x = (56 +/- 4i√6) / 8

factor

x = 4(14 +/- i√6) / 8

NOTE that 4 goes into 4 once and into 8 twice

x = 1(14 +/- i√6) / 2

x = (14 +/- i√6) / 2




you should be able to go from here.

OMG, just use Quadratic formula

1) x1= 7.7417 x2= 0.25834
2) x1= 1.4495 x2= -3.4495
3) x1= -2 plus 1.633 i x2= -2 minus 1.633 i
4) x1= 0 x2= -6
5) there is 4x²-56b+80 or 4x²+24 ???
6) x1= -2.5 0 x2= 2

hope this helps !!!