SOLUTION: The solutions to 2x^2=5x+7

Question 196128: The solutions to 2x^2=5x+7
Answer by bsmath(21) (Show Source): You can put this solution on YOUR website!
2x^2=5x+7 First, we need to put this quadratic equation in standard form:
subtract 5x+7 from both sides: 2x^2-5x-7=0. We use the quadratic equation because it cannot be easily factored. The quadratic equation is x=
-b/2a+-((b^2-4ac)^1/2/2a), where a is the coefficient of the x^2 term, b is the
coefficient of the x term, and c is the integral term. We have a=2, b=-5, and c=-7. Plugging in the values, x=5/2*2+-(25-4*2*-7)^1/2/2*2=5/4+-(25+56)^1/2/4,
x=(5+(81)^1/2)/4=14/4=7/2, x=(5-(81)^1/2)/4=-4/4=-1. Let's plug in x=-1 to see if we have it correct: 2(-1)^2-5(-1)-7=2+5-7=0, it is correct, x=-1 is a root.
You can plug in x=7/2 to see if it is a root.

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