SOLUTION: Ok, I have been working on these two since 4 30 this morning and still can't figure them out. Could somebody help me before I go nuts? One is (z-6)^2-40=0 and 2x^2+7x+2=0 I keep g

Question 180847: Ok, I have been working on these two since 4 30 this morning and still can't figure them out. Could somebody help me before I go nuts?
One is (z-6)^2-40=0 and 2x^2+7x+2=0 I keep getting some odd ball numbers and have not a clue what to do with them. Thanks in advance, Judy
Found 2 solutions by stanbon, Fombitz:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
(z-6)^2-40=0
z^2-12z+36-40 =0
z^2 - 12z - 4 = 0
Use the quadratic formula to get:
z = [12 +- sqrt(12^2 - 4*1*-4)]/2
z = [12 +- sqrt(160)]/2
z = [12 +- 4sqrt(10)]/2
z = 6 +- 2sqrt(10)
==========================
2x^2+7x+2=0
Use the quadratic forula to get:
x = [-7 +- sqrt(7^2 -4*2*2)]/4
x = [-7 +- sqrt(33)]/4
x = (-7/4) +- (1/4)sqrt(33)
===============================
Cheers,
Stan H.

Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!

Approximately z=12.32, -0.32.

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Use the quadratic formula,

Approximately, x=-3.18,-0.31

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Are those the oddball numbers you were getting??
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