How do i solve these equations?

Question

7x^2 - 30x = -8

3x^3- 48x= 0

another one is writing the equation of the axis of symmetry and finding the coordinates of the vertex of the graph y= -x^2 - 2x - 4

Answer 1

gayle325

7x^2 - 30x = -8

7x² - 30x + 8 = 0

Quadratic

a=7  b=(-30)  c=8

x =[-b ±√b² -4ac]/2a

x =[-(-30) ±√900 -4(7)(8)]/2(7)

x = [30 ±√900 - 224]/14

x = [30 ±√676] /14

x = [30 ± 26]/14

x+ = 27/7 = 55. 71

x- = 2/7 = 0.286

3x^3- 48x= 0

3x(x² - 16) =

3x = 0

x = 0

(x² - 16)

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

x = -4

x = 4

============================

Ans:: x = 0 , -4 & 4

============================

y = -x² -2x - 4

if y =0

0 = x² + 2x + 4

a=-1  b=-2  c=-4

x = [ -b ±√b² - 4ac]/2a

x = [ 2 ±√4 - 4(4)(-1)]/2(-1)

x = [ 2 ±√12]/-2

x = [ 2 ± 3.46]/-2

x+ = [2+3.46]/-2

x+ = - 2.73

x- = [ 2 - 3.46]/-2

x- = 2.93

if x = 0

y= -x^2 - 2x - 4

y = 0 - 0 -4

y = -4

if y = 0

x= (x+) + (x-)

x = (-2.73) + 2.93

x = 2

Coordinates

y=0 then x = 2 →(2,0)

x=0 then y = -4→(0,-4)

hope this helps

Answer 2

first one: get them all on one side and use the quadratic formula to solve.

second one:

x(3x^2 - 48) = 0

when two things multiply to make a zero, one of them must equal zero, so set them both equal to it

x = 0,     3x^2 - 48 = 0

you will get three answers total

Answer 3

7x^2 - 30x +8 = 0

(7x - 2)(x - 4) = 0

x = 2/7 or 4

3x(x^2 - 16) = 0

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

x = 0, 4 or -4

x = -2

(-2,-4)