how do you solve these 2 equations using the quadratic formula?

[erika]

How do you solve the equation using the quadratic formula?
3x^2 -17=0

or
2y^2=-5y



Thanks for your help !!
Thanks for all the answers. It makes perfect sense now.

[xclash]

2y^2+5y=0
y(2y+5=0

y=0 or -5/2

3x^2-17=0
3x^2=17
x^2=17/3
x= sq rt 17/3

[sadsdas]

x= [ -b +/- sqr(b^2 - 4ac) ] / 2a
where:
ax^2 + bx + c = 0

3x^2 -17=0
a= 3
b= 0
c= -17

plug and chug:
x= 2.3805, -2.3805

2y^2=-5y
2y^2 - 5y = 0
a= 2
b= -5
c= 0

plug and chug:
y= 2.5, 0

[hummingbird]

The quadratic formula says:
[-b ± ?(b²-4ab)] / 2a

1)
3x² - 17 = 0

So in this case:
a = 3
b = 0
c = -17

Plug them into the quadratic equation:
[0 ± ?(0 - 4(3)(-17))] / 6

[±?(204)] / 6

[±2?51] / 6

[±?51] / 3 <===Answer




2)
First rearrange the equation:
2y² + 5y = 0

So in this case:
a = 2
b = 5
c = 0

Plug it into the quadratic equation:
[-5 ± ?(25 - 4(2)(0))] / 4

[-5 ± ?25] / 4

[-5 ± 5] / 4

So you have to equations to solve for. I'll do the plus one first:
[-5 + 5] / 4=
0 <===Answer #1

Now the minus one:
[-5 - 5] / 4=
-5/2 <===Answer #2

[frankwu]

for 3x^2 - 17 = 0

ax^2 + bx + c = 0

a = 3
b = 0
c = -17

x= (-b(+/-)(b^2-4ac)^1/2)/2a  (quadratic equation)

x = (0(+/-)(0^2-((4)(3)(-17))^1/2)/2*3

x = ((+/-)(204)^1/2)/6

x = (+/-)2*((51)^(1/2))/6

x = pos 2/6 * sqrt51 and x = neg 2/6 * sqrt51

for 2y^2 = -5y, rearrange to:

2y^2+5y+0= 0

a=2
b=5
c=0

y = (-5(+/-)(5*5-4*2*0)^(1/2))/(2*2)

y = (-5(+/-)(25)^(1/2))/4

y = (-5 + 5)/4 and (-5 -5)/4

y = 0 and -10/4