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Q: How do you solve 2x2 plus 1x-3 equals 0?

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2x2-7x+3 = 0 (2x-1)(x-3) = 0 x = 1/2 or x = 3

The solutions work out as: x = 52/11, y = 101/11 and x = -2, y = -11

Differentiate the function with respect to x: d/dx (x3 - 2x2 - 5x + 6) = 3x2 - 4x - 5 Set this derivative = 0 and solve. 3x2 - 4x - 5 = 0 implies that x = -0.7863 or 2.1196 (to 4 dp)

-2x3 + 2x2 + 12x =(-2x) (x2 - x - 6) =(-2x) (x+2) (x-3)

2x2-5 = 0 +5 +5 2x2 = 5 /2 /2 x2 = 5/2 sqrt(x2) = sqrt (5/2) x = sqrt(5/2) Ps: Sqrt means square root

Related questions

-2x2 + 9x - 12 = 0Then apply the quadratic formula.

2x2 equals 5

To solve for x: 2x2 + 10 = 58 2x2 = 48 x2 = 24 x = √24 x = √4 × √6 x = 2√6

I'm sorry but I can't solve that problem. B(

3

There is no rational factorisation to this expression.

By using the quadratic equation formula: x = -5 and x = 3

2x2 + 11x + 15 = (x + 3)(2x + 5).

8 - 2x2 = x + 22 - 7Combine like terms on the right side:8 - 2x2 = x + 15Subtract 8 from each side:-2x2 = x + 7Add 2x2 to each side:0 = 2x2 + x + 7

-7

x2 + x2 = 2x2

No.

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