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

The solutions to the equation 2x^2-x=4 are x=(1+sqrt(33))/4 ,x=(1-sqrt(33))/4

Find the zeros of 2x^2-x=4

The zeros of 2x^2-x=4 are x=(1+sqrt(33))/4 ,x=(1-sqrt(33))/4

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2 x^{2} - x = 4

x = \frac{1 + 33}{4}, x = \frac{1 - 33}{4}

Decimal Notation

x = 1.68614 \dots, x = - 1.18614 \dots

Solution steps

2 x^{2} - x = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 2 (- 4) = 33

x_{1, 2} = \frac{- ( - 1 ) \pm 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 33}{2 \cdot 2}, x_{2} = \frac{- ( - 1 ) - 33}{2 \cdot 2}

x = \frac{- ( - 1 ) + 33}{2 \cdot 2} : \frac{1 + 33}{4}

x = \frac{- ( - 1 ) - 33}{2 \cdot 2} : \frac{1 - 33}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 33}{4}, x = \frac{1 - 33}{4}

25 x^{2} + 60 x + 37 = 0

x, 2 x^{2} = y^{2}

x^{2} - 5 x = 0

b (b - 15) = 30 - 2 b

(x - 9)^{2} = - 1

What are the solutions to the equation 2x^2-x=4 ?
The solutions to the equation 2x^2-x=4 are x=(1+sqrt(33))/4 ,x=(1-sqrt(33))/4
Find the zeros of 2x^2-x=4
The zeros of 2x^2-x=4 are x=(1+sqrt(33))/4 ,x=(1-sqrt(33))/4