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Do you mean 3x = 2 - 2x^2 ?

3x = 2 - 2x^2

3x - 2 + 2x^2 = 2 - 2x^2 -2 + 2x^2

2x^2 + 3x - 2 = 0

a = 2, b = 3, c = -2 (a, b, and c are real numbers)

The quadratic formula is:

x = [-b + & - square root of(b^2 - 4ac)]/2a, so

x = [-3 + & - square root of(3^2 - (4)(2)(-2)]/(2)(2)

x = [-3 + & - square root of(9 + 16)]/4

x = (-3 + & - square root of 25)/4

x = (-3 + & - 5)/4

x = (-3 + 5)/4 = 2/4 = 1/2 or

x = (-3 - 5)/4 = -8/4 = -2

The roots of the equation are -2 and 1/2.

The quantity that appears beneath the radical sign in the quadratic formula, b^2 - 4ac, is called the discriminant, which can tell you whether the roots of quadratic equation are real or imaginary.

If b^2 - 4ac = 0, there is one real root, called the double root.

If b^2 - 4ac > 0, there are two different real roots.

If b^2 - 4ac < 0, there are two conjugate imaginary roots.

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Q: Solve 3x2-2x2 using the quadratic formula Exopress your answer in simplest radical form?
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The is the name of the number under the radical symbol in the quadratic formula?

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How do you factor the trinomial x2 plus 4x minus 10?

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How do you solve complex cases of quadratic equations?

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What is an easy way to go the quadratic formula?

the quadratic formula is really complex. so, if you can avoid it, you should. one way i do this is by getting the trinomial into X^2 - 2YX + Y^2 = 0 form. here's an example lets say we have x^2 - 2x - 5 = 0 normally we would have to do the quadratic formula however, if we get it into this form, we can get x^2 - 2(1)x + 1 = 6 all i did was add 6 to both sides of the equation. you can now simplify the left side of the equation, and be left with (x-1)^(2) = 6 to get ride of the exponent on the left side, find the square root of both sides. this will leave you with (x-1) = (+ or -)radical(6) add one to both sides and you will get x = 1 (+ or -)radical(6) therefore, x can equal 1 + radical(6) or 1 - radical(6) now, this may seem complex, but it is easy once you get used to it. It is far more efficient than the long quadratic formula.