What are the solutions to the equation 1=2x^2+5x ?

The solutions to the equation 1=2x^2+5x are x=(-5+sqrt(33))/4 ,x=(-5-sqrt(33))/4

Find the zeros of 1=2x^2+5x

The zeros of 1=2x^2+5x are x=(-5+sqrt(33))/4 ,x=(-5-sqrt(33))/4

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1 = 2 x^{2} + 5 x

x = \frac{- 5 + 33}{4}, x = \frac{- 5 - 33}{4}

Decimal Notation

x = 0.18614 \dots, x = - 2.68614 \dots

Solution steps

1 = 2 x^{2} + 5 x

Switch sides

2 x^{2} + 5 x = 1

Move

1

to the left side

2 x^{2} + 5 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 1) = 33

x_{1, 2} = \frac{- 5 \pm 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 33}{2 \cdot 2}, x_{2} = \frac{- 5 - 33}{2 \cdot 2}

x = \frac{- 5 + 33}{2 \cdot 2} : \frac{- 5 + 33}{4}

x = \frac{- 5 - 33}{2 \cdot 2} : \frac{- 5 - 33}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 33}{4}, x = \frac{- 5 - 33}{4}

\frac{( x \cdot ( \frac{x}{4} + 1 ) )}{2} = 60

\frac{x ^{2}}{4} + 3 = 2 (x - 2)

2 a^{2} - 7 a + 3 = 0

2 = x^{2} - 12 x + 32

2 x^{2} + 3 x - 3 = 0

What are the solutions to the equation 1=2x^2+5x ?
The solutions to the equation 1=2x^2+5x are x=(-5+sqrt(33))/4 ,x=(-5-sqrt(33))/4
Find the zeros of 1=2x^2+5x
The zeros of 1=2x^2+5x are x=(-5+sqrt(33))/4 ,x=(-5-sqrt(33))/4