To know the equation with the solutions 2-3i and 2+3i
We will solve each using the quadratic formula;
![x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}](https://img.qammunity.org/2023/formulas/mathematics/college/rxvf73usjbbwyik14knxdemoz21vfz2ufc.png)
x^2 - 4x – 13=0
a=1 b=-4 c=-13
substitute the values above into the formula and then evaluate
![x=\frac{4\pm\sqrt[]{(-4)^2-4(1)(-13)}}{2(1)}](https://img.qammunity.org/2023/formulas/mathematics/college/zp14ag0q16aq6jp15upgtvgg4kf8oa0ymd.png)
![=\frac{4\pm\sqrt[]{16+52}}{2}](https://img.qammunity.org/2023/formulas/mathematics/college/vi4c6in799szi48vpvsrbpvi5zpdhun9i2.png)
![=(4)/(2)\pm\frac{\sqrt[]{68}}{2}](https://img.qammunity.org/2023/formulas/mathematics/college/xj14o1me5xn894qycw6hcebrqyyc8swyx3.png)
The above has no such root
Let's move on and check the second option
x^2 - 4x + 5=0
a=1 b= -4 c=5
substitute into the formula and evaluate
![x=\frac{4\pm\sqrt[]{(-4)^2-4(1)(5)}}{2}](https://img.qammunity.org/2023/formulas/mathematics/college/nbruwfnvqb9rj8bg98xzwmublpskzi8sen.png)
![=(4)/(2)\pm\frac{\sqrt[]{-4}}{2}](https://img.qammunity.org/2023/formulas/mathematics/college/fctwjfeeo5kyrbetudy3qy99dycm45347n.png)
x =2 ± i
The above is not the option
Let's check the next option
x^2 - 4x - 5
a=1 b= -4 c= -5
Option c is also not an option since c=-5, it will give a positive number on the square root
Let;s move on and check option D
x^2 - 4x + 13
a=1 b= -4 c=13
substitute into the formula and then evaluate
![x=\frac{4\pm\sqrt[]{(-4)^2-4(1)(13)}}{2}](https://img.qammunity.org/2023/formulas/mathematics/college/dpkhalhc6br4xv07vwslta9fe64b19acu1.png)
![x=(4)/(2)\pm\frac{\sqrt[]{-36}}{2}](https://img.qammunity.org/2023/formulas/mathematics/college/6443uj9an3rj8fa0z60d2b0pr9303grefh.png)
x= 2 ± 3i
Either x= 2-3i or x= 2+3i
The correct option is D
D. 0 = x^2 - 4x + 13