What are the solutions to the equation (2(-3/2)-1)x^2+4x+2=0 ?

The solutions to the equation (2(-3/2)-1)x^2+4x+2=0 are x=-(-1+sqrt(3))/2 ,x=(1+sqrt(3))/2

Find the zeros of (2(-3/2)-1)x^2+4x+2=0

The zeros of (2(-3/2)-1)x^2+4x+2=0 are x=-(-1+sqrt(3))/2 ,x=(1+sqrt(3))/2

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(2 (- \frac{3}{2}) - 1) x^{2} + 4 x + 2 = 0

x = - \frac{- 1 + 3}{2}, x = \frac{1 + 3}{2}

Decimal Notation

x = - 0.36602 \dots, x = 1.36602 \dots

Solution steps

(2 (- \frac{3}{2}) - 1) x^{2} + 4 x + 2 = 0

Expand

(2 (- \frac{3}{2}) - 1) x^{2} + 4 x + 2 : - 4 x^{2} + 4 x + 2

- 4 x^{2} + 4 x + 2 = 0

Solve with the quadratic formula

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

4^{2} - 4 (- 4) \cdot 2 = 43

x_{1, 2} = \frac{- 4 \pm 4 3}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 4 + 4 3}{2 ( - 4 )}, x_{2} = \frac{- 4 - 4 3}{2 ( - 4 )}

x = \frac{- 4 + 4 3}{2 ( - 4 )} : - \frac{- 1 + 3}{2}

x = \frac{- 4 - 4 3}{2 ( - 4 )} : \frac{1 + 3}{2}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 3}{2}, x = \frac{1 + 3}{2}

8 x^{2} + 14 x + 9 = 0

x^{2} + 24 + 144 = 0

x^{2} + x = 3 + 2 x

4 (12) = 3 x (x)

x, x^{2} - k x + 72 = 0

What are the solutions to the equation (2(-3/2)-1)x^2+4x+2=0 ?
The solutions to the equation (2(-3/2)-1)x^2+4x+2=0 are x=-(-1+sqrt(3))/2 ,x=(1+sqrt(3))/2
Find the zeros of (2(-3/2)-1)x^2+4x+2=0
The zeros of (2(-3/2)-1)x^2+4x+2=0 are x=-(-1+sqrt(3))/2 ,x=(1+sqrt(3))/2