What are the solutions to the equation 2x^2-x-29=15 ?

The solutions to the equation 2x^2-x-29=15 are x=(1+sqrt(353))/4 ,x=(1-sqrt(353))/4

Find the zeros of 2x^2-x-29=15

The zeros of 2x^2-x-29=15 are x=(1+sqrt(353))/4 ,x=(1-sqrt(353))/4

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2 x^{2} - x - 29 = 15

x = \frac{1 + 353}{4}, x = \frac{1 - 353}{4}

Decimal Notation

x = 4.94707 \dots, x = - 4.44707 \dots

Solution steps

2 x^{2} - x - 29 = 15

Move

15

to the left side

2 x^{2} - x - 44 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 ( - 44 )}{2 \cdot 2}

(- 1)^{2} - 4 \cdot 2 (- 44) = 353

x_{1, 2} = \frac{- ( - 1 ) \pm 353}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 353}{2 \cdot 2}, x_{2} = \frac{- ( - 1 ) - 353}{2 \cdot 2}

x = \frac{- ( - 1 ) + 353}{2 \cdot 2} : \frac{1 + 353}{4}

x = \frac{- ( - 1 ) - 353}{2 \cdot 2} : \frac{1 - 353}{4}

The solutions to the quadratic equation are:

x = \frac{1 + 353}{4}, x = \frac{1 - 353}{4}

3 x^{2} - 75 = 0

4 x^{2} + 22 x = 0

x^{2} = 16

x^{2} + 12 x = - 2 x - 40

x, - 5 k x^{2} + 8 x = 4 (x + 2)

What are the solutions to the equation 2x^2-x-29=15 ?
The solutions to the equation 2x^2-x-29=15 are x=(1+sqrt(353))/4 ,x=(1-sqrt(353))/4
Find the zeros of 2x^2-x-29=15
The zeros of 2x^2-x-29=15 are x=(1+sqrt(353))/4 ,x=(1-sqrt(353))/4