What are the solutions to the equation 2u^2+u-3=0 ?

The solutions to the equation 2u^2+u-3=0 are u=1,u=-3/2

Find the zeros of 2u^2+u-3=0

The zeros of 2u^2+u-3=0 are u=1,u=-3/2

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2 u^{2} + u - 3 = 0

u = 1, u = - \frac{3}{2}

Decimal Notation

u = 1, u = - 1.5

Solution steps

2 u^{2} + u - 3 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 3) = 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 1 + 5}{2 \cdot 2}, u_{2} = \frac{- 1 - 5}{2 \cdot 2}

u = \frac{- 1 + 5}{2 \cdot 2} : 1

u = \frac{- 1 - 5}{2 \cdot 2} : - \frac{3}{2}

The solutions to the quadratic equation are:

u = 1, u = - \frac{3}{2}

4 x^{2} = 36

\frac{x}{x - 1} - \frac{2}{x + 1} < \frac{8}{x ^{2} - 1}

7 (x - 3) + (x - 1) (x + 1) = x^{(2)} - 5 (x + 2)

\frac{x}{3} < 2

3 \cdot \frac{5}{2}

What are the solutions to the equation 2u^2+u-3=0 ?
The solutions to the equation 2u^2+u-3=0 are u=1,u=-3/2
Find the zeros of 2u^2+u-3=0
The zeros of 2u^2+u-3=0 are u=1,u=-3/2