SOLUTION: How would I solve for "x" in "7x^2-12x+9=0"?

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Question 897929: How would I solve for "x" in "7x^2-12x+9=0"?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

7x%5E2-12x%2B9=0 Start with the given equation.


Notice that the quadratic 7x%5E2-12x%2B9 is in the form of Ax%5E2%2BBx%2BC where A=7, B=-12, and C=9


Let's use the quadratic formula to solve for "x":


x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


x+=+%28-%28-12%29+%2B-+sqrt%28+%28-12%29%5E2-4%287%29%289%29+%29%29%2F%282%287%29%29 Plug in A=7, B=-12, and C=9


x+=+%2812+%2B-+sqrt%28+%28-12%29%5E2-4%287%29%289%29+%29%29%2F%282%287%29%29 Negate -12 to get 12.


x+=+%2812+%2B-+sqrt%28+144-4%287%29%289%29+%29%29%2F%282%287%29%29 Square -12 to get 144.


x+=+%2812+%2B-+sqrt%28+144-252+%29%29%2F%282%287%29%29 Multiply 4%287%29%289%29 to get 252


x+=+%2812+%2B-+sqrt%28+-108+%29%29%2F%282%287%29%29 Subtract 252 from 144 to get -108


x+=+%2812+%2B-+sqrt%28+-108+%29%29%2F%2814%29 Multiply 2 and 7 to get 14.


x+=+%2812+%2B-+6i%2Asqrt%283%29%29%2F%2814%29 Simplify the square root (note: If you need help with simplifying square roots, check out this solver)


x+=+%2812%2B6i%2Asqrt%283%29%29%2F%2814%29 or x+=+%2812-6i%2Asqrt%283%29%29%2F%2814%29 Break up the expression.


x+=+%282%286%2B3i%2Asqrt%283%29%29%29%2F%2814%29 or x+=+%282%286-3i%2Asqrt%283%29%29%29%2F%2814%29 Factor out the GCF


x+=+%286%2B3i%2Asqrt%283%29%29%2F%287%29 or x+=+%286-3i%2Asqrt%283%29%29%2F%287%29 Reduce


So the solutions are x+=+%286%2B3i%2Asqrt%283%29%29%2F%287%29 or x+=+%286-3i%2Asqrt%283%29%29%2F%287%29 where i+=+sqrt%28-1%29