What are the solutions to the equation 2x^2+x-41=-15 ?

The solutions to the equation 2x^2+x-41=-15 are x=(-1+sqrt(209))/4 ,x=(-1-sqrt(209))/4

Find the zeros of 2x^2+x-41=-15

The zeros of 2x^2+x-41=-15 are x=(-1+sqrt(209))/4 ,x=(-1-sqrt(209))/4

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2 x^{2} + x - 41 = - 15

x = \frac{- 1 + 209}{4}, x = \frac{- 1 - 209}{4}

Decimal Notation

x = 3.36420 \dots, x = - 3.86420 \dots

Solution steps

2 x^{2} + x - 41 = - 15

Move

15

to the left side

2 x^{2} + x - 26 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 26 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 26) = 209

x_{1, 2} = \frac{- 1 \pm 209}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 1 + 209}{2 \cdot 2}, x_{2} = \frac{- 1 - 209}{2 \cdot 2}

x = \frac{- 1 + 209}{2 \cdot 2} : \frac{- 1 + 209}{4}

x = \frac{- 1 - 209}{2 \cdot 2} : \frac{- 1 - 209}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 209}{4}, x = \frac{- 1 - 209}{4}

∣ 2 x - 15 ∣ \leq 7

4 x + 10 = 2 (2 x + 5)

65 \leq - 20 w - 11

\frac{1021}{26 e - 6 \cdot 0.1}

∣ 8 x ∣ = 5

What are the solutions to the equation 2x^2+x-41=-15 ?
The solutions to the equation 2x^2+x-41=-15 are x=(-1+sqrt(209))/4 ,x=(-1-sqrt(209))/4
Find the zeros of 2x^2+x-41=-15
The zeros of 2x^2+x-41=-15 are x=(-1+sqrt(209))/4 ,x=(-1-sqrt(209))/4