SOLUTION: hello :) hope anyone can help me Solve by using the quadratic equation X=- b+_ sqrt ( (B)^2 -4(a) (c) All over 2(a) 1.) a^2+2(a+3)=0 2.) x^2-3(2x+3)=0 3.) 3y(y-5)=6

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Question 980897: hello :) hope anyone can help me
Solve by using the quadratic equation
X=- b+_ sqrt ( (B)^2 -4(a) (c)
All over 2(a)
1.) a^2+2(a+3)=0
2.) x^2-3(2x+3)=0
3.) 3y(y-5)=6
4.) 2(3x^2+1) =
Hope anyone can help me solve :)
Kindly could you explain how too?
Thnak you bery much!

Found 2 solutions by Alan3354, MathLover1:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
2.) x^2-3(2x+3)=0
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Put in this format:
x^2-6x - 9 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=72 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 7.24264068711928, -1.24264068711928. Here's your graph:

---------------
Do the others the same way.
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#4 is not an equation, nothing on the right side.
Answer by MathLover1(20849)   (Show Source): You can put this solution on YOUR website!

1.)



...note your coefficient , unknown variable is , coefficient , and constant










solutions: there is no real solutions (so, there is no x-intercepts), just complex solutions


or






2.)










solutions:
exact

or


approximate:

or


3.)








solutions:




or




4.)






solutions: complex
=>=>

or
=>=>



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