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

Algebra ->  Algebra  -> Radicals -> 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      Log On

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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(48535) About Me  (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(13823) About Me  (Show Source):
You can put this solution on YOUR website!
%28z-6%29%5E2-40=0
%28z-6%29%5E2=40
z-6+=+0+%2B-+sqrt%2840%29+
z=6+%2B-+2%2Asqrt%2810%29
Approximately z=12.32, -0.32.
+graph%28+300%2C+300%2C+-4%2C+16%2C+-20%2C+20%2C+%28x-6%29%5E2-40%29+
.
.
.
2x%5E2%2B7x%2B2=0
Use the quadratic formula,
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-7+%2B-+sqrt%28+7%5E2-4%2A2%2A2+%29%29%2F%282%2A2%29+
x+=+%28-7+%2B-+sqrt%28+49-16+%29%29%2F%284%29+
x+=+%28-7+%2B-+sqrt%28+33+%29%29%2F%284%29+
Approximately, x=-3.18,-0.31
+graph%28+300%2C+300%2C+-7%2C+3%2C+-5%2C+5%2C+2%2Ax%5E2%2B7%2Ax%2B2%29+
.
.
.
Are those the oddball numbers you were getting??