How do you factor #2x^2+5x-6#?
1 Answer
Explanation:
Quadratic formula states that for an equation of the form
#x=(-b+sqrt(b^2-4ac))/(2a), (-b-sqrt(b^2-4ac))/(2a)# .
These two numbers are called roots.
Since plugging these 2 roots to
Thus, finding these roots means factoring the equation!
In this problem,
So
#x_1=(-5+sqrt(5^2-4(2)(-6)))/(2(2))# ,#x_2=(-5-sqrt(5^2-4(2)(-6)))/(2(2))# .
#x_1=(-5+sqrt(73))/(4)# ,#x_2=(-5-sqrt(73))/(4)# .
In conclusion,
#2x^2+5x-6=2(x+5/4-sqrt(73)/4)(x+5/4+sqrt(73)/4)# .
I multiplied the whole factored equation by 2 because the factored equation starts with