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=(3-sqrt(41))/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=(3-sqrt(41))/4

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

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

3 x^{3} - 5 x^{2} - 2 x

m^{2} = 17

1134831.46067 \cdot 16.04

4 x^{3} + 4 x^{2} - 7 x + 2

2 m^{2} - m - 6 = 0

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=(3-sqrt(41))/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=(3-sqrt(41))/4