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

The solutions to the equation 1=2x^2+7x are x=(-7+sqrt(57))/4 ,x=(-7-sqrt(57))/4

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

The zeros of 1=2x^2+7x are x=(-7+sqrt(57))/4 ,x=(-7-sqrt(57))/4

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1 = 2 x^{2} + 7 x

x = \frac{- 7 + 57}{4}, x = \frac{- 7 - 57}{4}

Decimal Notation

x = 0.13745 \dots, x = - 3.63745 \dots

Solution steps

1 = 2 x^{2} + 7 x

Switch sides

2 x^{2} + 7 x = 1

Move

1

to the left side

2 x^{2} + 7 x - 1 = 0

Solve with the quadratic formula

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

7^{2} - 4 \cdot 2 (- 1) = 57

x_{1, 2} = \frac{- 7 \pm 57}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 57}{2 \cdot 2}, x_{2} = \frac{- 7 - 57}{2 \cdot 2}

x = \frac{- 7 + 57}{2 \cdot 2} : \frac{- 7 + 57}{4}

x = \frac{- 7 - 57}{2 \cdot 2} : \frac{- 7 - 57}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 57}{4}, x = \frac{- 7 - 57}{4}

- 4 x^{2} + 6 x + 20 = 0

\frac{1}{2} x^{2} - 5 x = 0

a x + x y + y^{2} + a y

x^{2} + 20 = 4

3 x^{2} - 16 x + 15 = 3

What are the solutions to the equation 1=2x^2+7x ?
The solutions to the equation 1=2x^2+7x are x=(-7+sqrt(57))/4 ,x=(-7-sqrt(57))/4
Find the zeros of 1=2x^2+7x
The zeros of 1=2x^2+7x are x=(-7+sqrt(57))/4 ,x=(-7-sqrt(57))/4