Question

Solve the equation for all real solutions.
6p^2-5p-6=0

Answer 1

`\text{Use the quadratic formula:}\\\\\\p=(-b\pm√(b^2-4ac))/(2a)\\\\\text{Plug in and solve:}\\\\p=(-(-5)\pm√(-5^2-4(6)(-6)))/(2(6))\\\\p=(5\pm√(25+144))/(12)\\\\p=(5\pm√(169))/(12)\\\\p=(5\pm13)/(12)\\\\\text{Solve for addition:}\\\\p=(5+13)/(12)\\\\p=(18)/(12)\\\\\boxed{p=(3)/(2)}\\\\\text{Solve for subtraction:}\\\\p=(5-13)/(12)\\\\p=(-8)/(12)\\\\\boxed{p=-(2)/(3)}`

Answer 2

p= -2/3 or p=3/2

Explanation:

(3p +2)(2p -3)=0

3p=-2 or 2p=3

p=-2/3 or p=3/2