What are the solutions to the equation -16+2x^2+7x=1 ?

The solutions to the equation -16+2x^2+7x=1 are x=(-7+sqrt(185))/4 ,x=(-7-sqrt(185))/4

Find the zeros of -16+2x^2+7x=1

The zeros of -16+2x^2+7x=1 are x=(-7+sqrt(185))/4 ,x=(-7-sqrt(185))/4

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- 16 + 2 x^{2} + 7 x = 1

x = \frac{- 7 + 185}{4}, x = \frac{- 7 - 185}{4}

Decimal Notation

x = 1.65036 \dots, x = - 5.15036 \dots

Solution steps

- 16 + 2 x^{2} + 7 x = 1

Move

1

to the left side

2 x^{2} + 7 x - 17 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 ( - 17 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 17) = 185

x_{1, 2} = \frac{- 7 \pm 185}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 185}{2 \cdot 2}, x_{2} = \frac{- 7 - 185}{2 \cdot 2}

x = \frac{- 7 + 185}{2 \cdot 2} : \frac{- 7 + 185}{4}

x = \frac{- 7 - 185}{2 \cdot 2} : \frac{- 7 - 185}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 185}{4}, x = \frac{- 7 - 185}{4}

3 X^{2} - 12 = 0

25 x^{2} - 50 x + 9 = 0

9 - x^{2} = 12

y^{2} + 6 y - 112 = 0

(1 - x) (3 - x) = 8

What are the solutions to the equation -16+2x^2+7x=1 ?
The solutions to the equation -16+2x^2+7x=1 are x=(-7+sqrt(185))/4 ,x=(-7-sqrt(185))/4
Find the zeros of -16+2x^2+7x=1
The zeros of -16+2x^2+7x=1 are x=(-7+sqrt(185))/4 ,x=(-7-sqrt(185))/4