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

The solutions to the equation -2x^2-3x+4=0 are x=-(3+sqrt(41))/4 ,x=(sqrt(41)-3)/4

Find the zeros of -2x^2-3x+4=0

The zeros of -2x^2-3x+4=0 are x=-(3+sqrt(41))/4 ,x=(sqrt(41)-3)/4

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- 2 x^{2} - 3 x + 4 = 0

x = - \frac{3 + 41}{4}, x = \frac{41 - 3}{4}

Decimal Notation

x = - 2.35078 \dots, x = 0.85078 \dots

Solution steps

- 2 x^{2} - 3 x + 4 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 (- 2) \cdot 4 = 41

x_{1, 2} = \frac{- ( - 3 ) \pm 41}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 41}{2 ( - 2 )}, x_{2} = \frac{- ( - 3 ) - 41}{2 ( - 2 )}

x = \frac{- ( - 3 ) + 41}{2 ( - 2 )} : - \frac{3 + 41}{4}

x = \frac{- ( - 3 ) - 41}{2 ( - 2 )} : \frac{41 - 3}{4}

The solutions to the quadratic equation are:

x = - \frac{3 + 41}{4}, x = \frac{41 - 3}{4}

- 72 + 6 w + w^{2}

a^{2} - a - 20 = 0

1 + x \geq 2 x

8^{2} - 2^{5} = x

0 = 205 - 11 t - 16 t^{2}

What are the solutions to the equation -2x^2-3x+4=0 ?
The solutions to the equation -2x^2-3x+4=0 are x=-(3+sqrt(41))/4 ,x=(sqrt(41)-3)/4
Find the zeros of -2x^2-3x+4=0
The zeros of -2x^2-3x+4=0 are x=-(3+sqrt(41))/4 ,x=(sqrt(41)-3)/4