Respuesta :
quadratic formula where:
[tex]ax ^{2} + bx + c[/tex]
[tex] \frac{ - b + { \sqrt{b^{2} - 4ac } }}{2a} [/tex]
[tex] \frac{ - b - \sqrt{b { }^{2} - 4ac } }{2a} [/tex]
do both do get the roots of the equation
[tex]ax ^{2} + bx + c[/tex]
[tex] \frac{ - b + { \sqrt{b^{2} - 4ac } }}{2a} [/tex]
[tex] \frac{ - b - \sqrt{b { }^{2} - 4ac } }{2a} [/tex]
do both do get the roots of the equation
Answer:
Step-by-step explanation:
A short answer is that you cannot factor this the way you are trying to do it. What to do to go on is a question. You could use the quadratic formula to get some sort of answer but if factoring is your only option, forget about it.
Here's how you can check.
- a = 12
- b = 1
- c = -15
There is a part of the quadratic solution is the sqrt(b^2 - 4*a*c). In order to be factorable, this sqrt must result in a perfect square.
sqrt(1 - 4(12)(-15) ) = sqrt( 1 - - 720) = sqrt(721)
This is anything but a perfect square. (721 does factor, however, into 7 * 103). I03 is prime. So the long and short of it is, there's no way it will factor.