Answer:
Two imaginary solutions:
x₁= 
x₂ = 
Step-by-step explanation:
When we are given a quadratic equation of the form ax² +bx + c = 0, the discriminant is given by the formula b² - 4ac.
The discriminant gives us information on how the solutions of the equations will be.
- <u>If the discriminant is zero</u>, the equation will have only one solution and it will be real
- <u>If the discriminant is greater than zero</u>, then the equation will have two solutions and they both will be real.
- <u>If the discriminant is less than zero,</u> then the equation will have two imaginary solutions (in the complex numbers)
So now we will work with the equation given: 4x - 3x² = 10
First we will order the terms to make it look like a quadratic equation ax²+bx + c = 0
So:
4x - 3x² = 10
-3x² + 4x - 10 = 0 will be our equation
with this information we have that a = -3 b = 4 c = -10
And we will find the discriminant: 
Therefore our discriminant is less than zero and we know<u> that our equation will have two solutions in the complex numbers. </u>
To proceed to solve the equation we will use the general formula
x₁= (-b+√b²-4ac)/2a
so x₁ = 
The second solution x₂ = (-b-√b²-4ac)/2a
so x₂=
These are our two solutions in the imaginary numbers.