Solve by factoringg, and a word problem?

Question

this is freaking c**p.

2y^2 + 7y + 3= 0

4w^2 = 4w + 3

6x^2 + 5x = 4

ohh && also if someone cud help me with this one;

use formula h= -16t^2 + Vo2

a ball is thrown upward at an initial speed of 40ft/s

a. when does the ball reach a height of 24ft?

b. When does it reach a height of 48ft?

c. what is the greatest height reached by the ball?

d. when does the ball hit the ground?

thanksss.

Answer 1

Factoring

2y^2+7y+3=0

(2y+1)(y+3)=0

2y+1=0     y+3=0

    -1  -1

2y= -1

y= -1/2

4w^2 - 4w - 3=0

(2w+1)(2w-3)=0

2w+1=0   2w-3=0

w= -1/2     w= 3/2

6x^2+5x-4=0

(3x+4)(2x-1)=0

3x+4=0   2x-1=0

x= -4/3    x= 1/2

Answer 2

2y^2+7y+3=0

y^2+7y+6=0

(y+6)(y+1)=0

Divide 6 and 1 by 2, and simplify. Bring the denominator in front of the y.

(y+6/2)(y+1/2)=0

(y+3)(2y+1)=0

4w^2-4w-3=0

w^2-4w-12=0

(w-6)(w+2)=0

(w-6/4)(w+2/4)=0

(w-3/2)(w+1/2)=0

(2w-3)(2w+1)=0

6x^2+5x-4=0

x^2+5x-24=0

(x-3)(x+8)=0

(x-3/6)(x+8/6)=0

(x-1/2)(x+4/3)=0

(2x-1)(3x+4)=0

Vo=initial velocity

h=-16t^2+40t

a)24=-16t^2+40t

Divide by 8

3=-2t^2+5t

2t^2-5t+3=0

t^2-5t+6=0

(t-2)(t-3)=0

(t-2/2)(t-3/2)=0

t=1 second

t=3/2 second

b. Same thing as a.

c. -b/2a=x coordinate of vertex

-40/-32=5/4. Plug that in to -16t^s+40t and thats the greatest height.

d. -16t^2+40t=0

-8t(2t-5)=0

t=0

t=5/2