How would I work these, I'm lost...HELP!?

Question

4x^2 + 10x + 3 = 0

x^2 = 3 - 4x

-4x^2 - 3x - 1 = 0

(3x + 5)^2 = 100

I know I'm suppose to use the quadratic formula for the first one (which I did, but I came up with a different answer than the choices given) but I don't know how to work the others. HELP!

Answer 1

To use the quadratic formula, the equation must be in ax^2 + bx + c = 0 form.

x = (-b +/- sqrt (b^2 - 4ac)) / (2a)
so for #1,
x = (-10 +/- sqrt (10^2 - 4*4*3)) / (2*4)
= (-10 +/- sqrt (100-48) ) / 8
= (-10 +/- sqrt 52) / 8
= (-10 +/- 2 sqrt 13) / 8
= (-5 /- sqrt 13) / 4

Answer 2

The quadratic equation is ax^2+bx+c=0
There are 2 solutions.
x=[-b+sqrt(b^2-4ac)]/2a
x=[-b-sqrt(b^2-4ac)]/2a
4x^2+10x+3=0
a=4 b=10 c=3
x=[-10+sqrt(100-48)]/8
=[-10+7.874]/8
the other x=[-10-7.874]/8
Use the same procedure for other problems.
x^2=3-4x
x^2+4x-3=0
a=1 b=4 c=-3

(3x+5)^2=100
9x^2+30x+25=100
9x^2+30x-75=0
a=9 b=30 c=-75

Answer 3

4x^2 + 10x + 3 = 0
x = [-10 +/- sqrt(10^2-4(4)(3))]/(2*4)
x = [-10 +/- sqrt(52)]/8
x = [-10 +/- 2sqrt(13)]/8
x = [-5 +/- sqrt(13)]/4

x^2 = 3 - 4x
x^2 +4x -3 = 0
x = [-4 +/- sqrt(4^2-4(1)(3))]/2
x = [-4 +/- sqrt(4)]/2
x = -2 +/- 2

-4x^2 - 3x - 1 = 0
x = [3 +/- sqrt(3^2 -4(-4)(-1)]/(2* -4)
x = [3 +/- sqrt(-7))]/-8
x = [3 +/- isqrt(7)]/-8

(3x + 5)^2 = 100
3x+5 = +/- 10
3x = -5 +/- 10
x = -5(1 +/-2 )/3

Answer 4

Either factor or use the quadratic formula on all of the equations. However, you will need to rearrange the second thru fourth to look like the first.

You can do it....Sure, I knew you could.