SOLUTION: Could I get someone to show me how to solve this equation correctly? Use the quadratic formula to solve the following quadratic equation. {{{2x^2-5x=3}}}

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Question 100455: Could I get someone to show me how to solve this equation correctly?
Use the quadratic formula to solve the following quadratic equation.
2x%5E2-5x=3
Found 2 solutions by checkley71, edjones:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
2X^2-5X-3=0
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
X=(5+-SQRT[-5*2-4*2*-3])/2*2
Y=(5+-SQRT[25+24])/4
X=(5+-SQRT49)/4
X=(5+-7)/4
X=(5+7)/4
X=12/4
X=3 ANSWER.
X=(5-7)/4
X=-2/4
X=-1/2 ANSWER.

Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E2-5x=3
2x^2-5x-3=0 subtract 3 from each side
Ed
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-5x%2B-3+=+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=%28-5%29%5E2-4%2A2%2A-3=49.

Discriminant d=49 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+49+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+49+%29%29%2F2%5C2+=+3
x%5B2%5D+=+%28-%28-5%29-sqrt%28+49+%29%29%2F2%5C2+=+-0.5

Quadratic expression 2x%5E2%2B-5x%2B-3 can be factored:
2x%5E2%2B-5x%2B-3+=+%28x-3%29%2A%28x--0.5%29
Again, the answer is: 3, -0.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-5%2Ax%2B-3+%29