Helpppp.
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2007-11-15 09:29:20 · 5 answers · asked by Mariah M 1 in Science & Mathematics ➔ Mathematics
Well first, you have to bring 5x^2 over to the other side, to get the quadratic equation. (ax^2 +bx+c)
Now, you'll have -5x^2+3x-7=0.
a= -5
b=3
c=-7
If you only need the discriminant, then you only have to use the part of the quadratic formula that's underneath the radical sign.
x = [ -b ± sqrt(b^2 - 4ac) ] / 2a
That is the complete quadratic formula, but as I said you only need the part underneath the radical, which is radical b^2-4ac.
Now, you have to substitute your problem into the formula.
b^2 becomes 3^2
Then you have -4ac which becomes -4 (-5) (-7)
All together it's 3^2 -4(-5)(-7).
3^2 is 9
-4 times -5 times -7 = -140
9-140 = -131
-131 is your discriminant.
2007-11-15 09:37:58 · answer #1 · answered by Kate 3 · 0⤊ 0⤋
5x^2 - 3x + 7 = 0
discriminant = b^2 - 4*a*c
= (-3)^2 - 4*5*7
= 9 - 140 = - 131
2007-11-15 17:34:05 · answer #2 · answered by Zmik 3 · 0⤊ 0⤋
bring all members on one side of the "="
5x^2-3x+7=0
now you have the form:
ax^2+bx+c=0
then the discriminant is: b^2-4ac
a= 5
b= -3
c= 7
then
(-3)^2-4*5*7 = 9 - 140 = -131
bye!!!
2007-11-15 17:34:45 · answer #3 · answered by Matteo Man 3 · 0⤊ 0⤋
5x^2-3x +7=0
b^2- 4ac = (-3)^2 -4(5)(7) = 9 - 140 = -131
2007-11-15 17:32:52 · answer #4 · answered by ironduke8159 7 · 1⤊ 0⤋
I'll be doing that stuff very soon
2007-11-15 17:31:37 · answer #5 · answered by calebrules1991 5 · 0⤊ 2⤋