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Mathematics 13 Online
OpenStudy (deercult):

Solve the equation using the quadratic formula. x^2 + 10x = -25

OpenStudy (candy13106):

ok so do u know the quadratic formula

OpenStudy (deercult):

I know the quadratic formula... for ax^2 + bx + c = 0, \[x = \frac{ -b \pm \sqrt{b ^{2}}-4ac }{ 2a }\]

OpenStudy (deercult):

Whoops. Sqrt the -4ac too though.

OpenStudy (candy13106):

ok so u would plug 1x for a 10x for b and 25 for c cause u have to equal it to 0

OpenStudy (candy13106):

first step is making it ax^2+bx+c=0

OpenStudy (candy13106):

if its -b+ and - u put (-10)

OpenStudy (candy13106):

no. its not a dumb question

OpenStudy (deercult):

Sorry, I saw you say 25 in place of c just now. I have /really/ bad number memory. And thank you. So is this it? \[\frac{ -10 \pm \sqrt{10^{2}-4(25)} }{ 2 }\] Since a is just 1

OpenStudy (candy13106):

yea so u would multiply -4 times 25 and 10^2

OpenStudy (candy13106):

so 100 - what ever u got when u multiplied

OpenStudy (candy13106):

u confused?

OpenStudy (deercult):

No, sorry! -4*25 is -100, so it's 100 - (-100) / 100 + 100= 200, yeah?

OpenStudy (candy13106):

ok so its actually 100-100 becuase u dont need to have the (-100) in parentheses

OpenStudy (deercult):

Ah, alright

OpenStudy (candy13106):

so square root of 0 is 0 and -10+the square root of 0 is -10/2

OpenStudy (candy13106):

and the do -10-the square root of 0 and divide by 2

OpenStudy (deercult):

So x = -5?

OpenStudy (candy13106):

and x= also another number

OpenStudy (deercult):

The only answer options are all one number, plus -10+0 and -10-0 would bring about the same results.

OpenStudy (candy13106):

true hold on let me do the problem on paper. i did this in my head and think i messed up

OpenStudy (deercult):

Alright, no problem.

OpenStudy (candy13106):

ok so i got -5

OpenStudy (deercult):

Ok! Plugging it into the original equation gets the right answer anyways, so I'm pretty confident that's correct. Thanks for your help!

OpenStudy (candy13106):

u welcome @ me if u need anymore help

OpenStudy (deercult):

Will do!

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