SOLUTION: find the real solutions, using the quadratic equation 3x(x+2)=5

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Question 304686: find the real solutions, using the quadratic equation
3x(x+2)=5
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
3x%28x%2B2%29=5
3x%5E2+%2B+6x+=+5
3x%5E2+%2B+6x+-+5+=+0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B6x%2B-5+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%286%29%5E2-4%2A3%2A-5=96.

Discriminant d=96 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-6%2B-sqrt%28+96+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%286%29%2Bsqrt%28+96+%29%29%2F2%5C3+=+0.632993161855452
x%5B2%5D+=+%28-%286%29-sqrt%28+96+%29%29%2F2%5C3+=+-2.63299316185545

Quadratic expression 3x%5E2%2B6x%2B-5 can be factored:
3x%5E2%2B6x%2B-5+=+3%28x-0.632993161855452%29%2A%28x--2.63299316185545%29
Again, the answer is: 0.632993161855452, -2.63299316185545. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B6%2Ax%2B-5+%29