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Mathematics 18 Online
OpenStudy (phebe):

Solve by quadratic formula 4x^2 + 3x – 8 = 0

OpenStudy (phebe):

@yahman

OpenStudy (anonymous):

Quadratic Equation Ax2 + Bx + C = 0

OpenStudy (phebe):

yes

OpenStudy (phebe):

no thas not the formula

OpenStudy (anonymous):

\[x=\frac{ -b \pm \sqrt{b^{2}-4ac} }{ 2a }\]

OpenStudy (phebe):

yes thas the formula

OpenStudy (anonymous):

ohh ok

OpenStudy (phebe):

@charliealbert

OpenStudy (anonymous):

Just substitute and you will get the answer

OpenStudy (phebe):

k

OpenStudy (anonymous):

a=4 b=3 c= -8

OpenStudy (phebe):

is that the answer

OpenStudy (anonymous):

No, just susbtitute thos values in the formula before

OpenStudy (phebe):

ok.......................

OpenStudy (phebe):

um...................... help mee

OpenStudy (anonymous):

In what?

OpenStudy (phebe):

its hardd

OpenStudy (anonymous):

TO subsitute?

OpenStudy (phebe):

noo

OpenStudy (phebe):

im jus in a hurry nd don't feel lik doin all that

OpenStudy (phebe):

lol lil lazzyy

OpenStudy (anonymous):

ok step 1 is to find the coefficients a, b, and c done ! a = 4 b = 3 c =-8

OpenStudy (phebe):

yepp i got that

OpenStudy (anonymous):

2nd step is plug in the values so .. \[\frac{ -3\pm \sqrt{3^{2}-4\times4\times}(-8) }{ 2\times4 }\]

OpenStudy (anonymous):

step 3 simplify expression under square root so .. \[\frac{ -3\pm \sqrt{137} }{ 8 }\]

OpenStudy (phebe):

ok

OpenStudy (phebe):

thas the answer

OpenStudy (anonymous):

no step 4 solve for x

OpenStudy (phebe):

ok

OpenStudy (anonymous):

\[\frac{ -3+\sqrt{137} }{ 8 }=-\frac{ 3 }{ 8 }+\frac{ 1 }{ 8 }\sqrt{137}\]

OpenStudy (anonymous):

^ x1

OpenStudy (anonymous):

\[\frac{ -3-\sqrt{137} }{ 8 }=-\frac{ 3 }{ 8 }-\frac{ 1 }{ 8 }\sqrt{137}\]

OpenStudy (anonymous):

^ x2

OpenStudy (anonymous):

My work here is done : )

OpenStudy (phebe):

so x^2 is the answer

OpenStudy (anonymous):

lol no omg

OpenStudy (phebe):

XDXD so lol wat was it

OpenStudy (phebe):

@Shay17

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