What are the solutions to the equation -2x^2-3x+12=0 ?

The solutions to the equation -2x^2-3x+12=0 are x=-(3+sqrt(105))/4 ,x=(sqrt(105)-3)/4

Find the zeros of -2x^2-3x+12=0

The zeros of -2x^2-3x+12=0 are x=-(3+sqrt(105))/4 ,x=(sqrt(105)-3)/4

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- 2 x^{2} - 3 x + 12 = 0

x = - \frac{3 + 105}{4}, x = \frac{105 - 3}{4}

Decimal Notation

x = - 3.31173 \dots, x = 1.81173 \dots

Solution steps

- 2 x^{2} - 3 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 2 ) \cdot 12}{2 ( - 2 )}

(- 3)^{2} - 4 (- 2) \cdot 12 = 105

x_{1, 2} = \frac{- ( - 3 ) \pm 105}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 105}{2 ( - 2 )}, x_{2} = \frac{- ( - 3 ) - 105}{2 ( - 2 )}

x = \frac{- ( - 3 ) + 105}{2 ( - 2 )} : - \frac{3 + 105}{4}

x = \frac{- ( - 3 ) - 105}{2 ( - 2 )} : \frac{105 - 3}{4}

The solutions to the quadratic equation are:

x = - \frac{3 + 105}{4}, x = \frac{105 - 3}{4}

(x - 4)^{2} = - 4

3 x^{2} - 4 x < 2 x^{2} + 21

\frac{240 - 80 x}{3} \cdot 0.22 = \frac{80 x}{3} \cdot 0.21

16 x^{2} = 56 x - 51

6 x (x - 4)^{4} + 12 x^{2} (x - 4)^{3}

What are the solutions to the equation -2x^2-3x+12=0 ?
The solutions to the equation -2x^2-3x+12=0 are x=-(3+sqrt(105))/4 ,x=(sqrt(105)-3)/4
Find the zeros of -2x^2-3x+12=0
The zeros of -2x^2-3x+12=0 are x=-(3+sqrt(105))/4 ,x=(sqrt(105)-3)/4