![(-b+-√(b^2-4ac) )/(2a); ax^2+bx+c \\ \\ a=1, b=-12, c=59](https://img.qammunity.org/2019/formulas/mathematics/middle-school/snfkg8s8zst7a6o8lmxkpkr982x0f088m1.png)
Use the formulas above to solve the equation.
![1x^2-12x+59=0\\ \\ (-(-12)+-√((-12)^2-4(1)(59)) )/(2(1))\\ \\ (12+-√(144-236) )/(2)\\ \\ (12+-√(-92) )/(2)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/hxq8iun6kh1w51s8z5a2j5evymn6loe99m.png)
Now this problem will be giving i values since there's a negative in the square root. These i's are known as imaginary numbers. For the answer it should be these two:
x =(12-
)/2=6-i
= 6.0000-4.7958i
x =(12+
)/2=6+i
= 6.0000+4.7958i
Let me know if you need help with finding i and all or if you just needed the answer to check your work. Hope this helps!