Question 531695: How do I solve for x in these equations?
-7=x^2-8x
10=2x^2+5x+13
Found 2 solutions by oberobic, KMST:
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! -7=x^2-8x
x^2-8x=-7
x^2-8x +7 = 0
(x-1)(x-7) = 0
x = 1 or 7
.
10 = 2x^2+5x+13
2x^2+5x+13=10
2x^2 +5x +3 = 0
(2x+3)(x+1) = 0
x = -1 or -3/2
Answer by KMST(5328)  (Show Source):
You can put this solution on YOUR website! In certain cases, you may ''see'' the solution, but there are 3 general strategies to solve quadratic equations: factoring, completing the square, and using the quadratic formula.
Factoring works often (but not always) for classroom problems.
For example,
can be written as
and since we can factor that quadratic trinomial as
the equation is equivalent to
 ,
and the product is zero when and only when one of the factors is zero, for
 or  , the solutions of the equation.
Another way of solving the equation would be "completing the square" (adding a number to both sides to make the one with the x into a perfect square).
so adding 16 to both sides of the first equation, you have
 which is equivalent to
so  or 
The same approach applied to a generic quadratic equation
 leads to the quadratic formula
As long as you can remember it (or read it), and you can apply it correctly, the quadratic formula will solve any quadratic equation. What I mean is that it will tell you the real solutions, if there is one or two of those, and it will tell you that there are no real solutions if there are none. If the expression under the square root (called the discriminant) is negative, there will be no real solutions.
For the second equation, i would transform it into
 and then I would apply the quadratic formula, because in that case factoring is not that easy, and neither is completing the square.
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