Answer:
The solutions are -7/2 and 1. Jeffrey is correct.
Explanation:
We have to use the Bhaskara formula to solve this question.
Bhaskara formula:
Suppose we have the following second order equation:
ax² + bx + c = 0
The solution of the equation are:
![x=(-b\pm√(b^2-4ac))/(2a)](https://img.qammunity.org/2023/formulas/mathematics/college/jr19ixi2zltkocy82qhxfiop5lyv4hzbkm.png)
In this question:
(2x - 1)(x + 3) = 4
We have to apply the distributive property to place the equation in the correct format to apply Bhaskara.
(2x - 1)(x + 3) = 4
2x² + 6x - x - 3 = 4
2x² + 5x - 3 - 4 = 0
2x² + 5x - 7 = 0
![x=(-5\pm√((5)^2-4\ast2\ast(-7)))/(2\ast2)=(-5\pm√(25+56))/(4)=(-5\pm√(81))/(4)=(-5\pm9)/(4)](https://img.qammunity.org/2023/formulas/mathematics/college/8ejjrkchxzty8fuhvw8csx408thh0x0an5.png)
The solutions are:
![x^{^(\prime)}=(-5+9)/(4)=1](https://img.qammunity.org/2023/formulas/mathematics/college/kwc9pfjq0b6tre89nnpadz468xkow29o73.png)
![x^{^(\prime\prime)}=(-5-9)/(4)=-(14)/(4)=-(7)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/dlvjjtwcgnz6dl8uu4xp1swhq2jo8sf4oq.png)
The solutions are -7/2 and 1. Jeffrey is correct.