Question 107049: i need help solving these quadratic equations using the quadratic formula.
2x^2 - 5x = 3
3x^2 - 2x + 1 = 0
i am trying to solve for x
2(x-5)^2 = 3
and this queston also i have been trying to figure these out fo an hour.
x^2 + 4x + 4 = 7
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! You must get them in the standard form of ax^2+bx+c=0.
2x^2 - 5x = 3
2x^2-5x-3=0
a=2, b=-5, c=-3
.
2(x-5)^2 = 3
2(x^2-10x+25)=3
2x^2-20x+50-3=0
2x^2-20x+47=0
.
x^2 + 4x + 4 = 7
x^2+4x-3=0
.
I'll do two and you can do the rest.
Ed
| 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=49 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: 3, -0.5.
Here's your graph:
 |
| 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=0 is zero! That means that there is only one solution:  .
Expression can be factored: 
Again, the answer is: 5, 5.
Here's your graph:
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