Answer:
Solving the expression
we get:
![\mathbf{x=(1+√(2)i )/(3)\:or\:x=(1-√(2)i )/(3)}](https://img.qammunity.org/2022/formulas/mathematics/college/g1ib6rmai8eqmzpbufygv8189cfm7hc7oe.png)
Explanation:
We need to solve the expression:
![3x^2-2x=-1](https://img.qammunity.org/2022/formulas/mathematics/college/ha5jwlqexiailkrp0kfk4pzb6k2sl088z9.png)
This is a quadratic expression and it can be solved using quadratic formula
Solving:
![3x^2-2x=-1\\](https://img.qammunity.org/2022/formulas/mathematics/college/w04zt850wyr1ouix8ia7s0xw3zxbqi5fo3.png)
we can write it as:
![3x^2-2x+1=0](https://img.qammunity.org/2022/formulas/mathematics/college/52i0inbcqcp15r2dvugwmsf7rjcvvs3ji4.png)
The quadratic formula is:
![x=(-b\pm√(b^2-4ac))/(2a)](https://img.qammunity.org/2022/formulas/mathematics/college/xwoste9sb5s8xgnw06frxmsrdev3dn4cxi.png)
where a = 3, b = -2 and c= 1
Putting values and solving:
![x=(-b\pm√(b^2-4ac))/(2a)\\x=(-(-2)\pm√((-2)^2-4(3)(1)))/(2(3))\\x=(2\pm√(4-12))/(2(3))\\x=(2\pm√(-8))/(6)\\We\:know\:that\:√(-1)=i\\x=(2\pm√(8)√(-1) )/(6) \\We\:know\:√(8)=√(2* 2 * 2)=√(2^2 * 2)=2√(2) \\x=(2\pm2√(2)i )/(6)\\Now,\\x=(2+2√(2)i )/(6)\:or\:x=(2-2√(2)i )/(6)\\x=(2(1+√(2)i) )/(6)\:or\:x=(2(1-√(2)i) )/(6)\\x=(1+√(2)i )/(3)\:or\:x=(1-√(2)i )/(3)](https://img.qammunity.org/2022/formulas/mathematics/college/kqay1eksg3q8j4uqjaq7dcgqy8dp0mhi6t.png)
So, solving the expression
we get:
![\mathbf{x=(1+√(2)i )/(3)\:or\:x=(1-√(2)i )/(3)}](https://img.qammunity.org/2022/formulas/mathematics/college/g1ib6rmai8eqmzpbufygv8189cfm7hc7oe.png)