SOLUTION: how do i solve 9h^2-6h+7=0 by using the quadratic formula

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Question 235514: how do i solve 9h^2-6h+7=0 by using the quadratic formula
Found 2 solutions by josmiceli, solver91311:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
9h%5E2+-+6h+%2B+7+=+0
h+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a,b and c in this formula are the coefficients
in the general formula:
ax%5E2+%2B+bx+%2B+c+=+0 (substitute h's for x's)
a+=+9
b+=+-6
c+=+7
h+=+%28-%28-6%29+%2B-+sqrt%28+%28-6%29%5E2-4%2A9%2A7+%29%29%2F%282%2A9%29+
h+=+%28+6+%2B-+sqrt%28+36+-+252+%29%29%2F18+
h+=+%28+6+%2B-+sqrt%28+-216+%29%29%2F18+
h+=+%28+6+%2B-+sqrt%28+-6%2A36+%29%29%2F18+
h+=+%28+6+%2B-+6%2Asqrt%28+-6+%29%29%2F18+
h+=+1%2F3+%2B+sqrt%286%29i%2F3
and
h+=+1%2F3+-+sqrt%286%29i%2F3
(i+=+sqrt%28-1%29)

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


If you have a quadratic equation of the form , then the two roots can be calculated using the coefficients of the quadratic trinomial thus:



In your case, the quadratic is in and the coefficients are , , and . Just plug in the numbers and do the arithmetic:



In this problem, the quantity under the radical, that is to say is less than zero. That means that your two roots will be a conjugate pair of complex roots of the form where is the imaginary number defined by

John