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

The solutions to the equation 2x^2+5x=2 are x=(-5+sqrt(41))/4 ,x=(-5-sqrt(41))/4

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

The zeros of 2x^2+5x=2 are x=(-5+sqrt(41))/4 ,x=(-5-sqrt(41))/4

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2 x^{2} + 5 x = 2

x = \frac{- 5 + 41}{4}, x = \frac{- 5 - 41}{4}

Decimal Notation

x = 0.35078 \dots, x = - 2.85078 \dots

Solution steps

2 x^{2} + 5 x = 2

Move

2

to the left side

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

Solve with the quadratic formula

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

5^{2} - 4 \cdot 2 (- 2) = 41

x_{1, 2} = \frac{- 5 \pm 41}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 41}{2 \cdot 2}, x_{2} = \frac{- 5 - 41}{2 \cdot 2}

x = \frac{- 5 + 41}{2 \cdot 2} : \frac{- 5 + 41}{4}

x = \frac{- 5 - 41}{2 \cdot 2} : \frac{- 5 - 41}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 41}{4}, x = \frac{- 5 - 41}{4}

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

3 y^{2} - 108 = 0

x, z = \frac{( x ^{2} - y ^{2} )}{2}

2 x^{2} + x = 4 x - 2

- 4.9 t^{2} + 11.43 + 15 = 0

What are the solutions to the equation 2x^2+5x=2 ?
The solutions to the equation 2x^2+5x=2 are x=(-5+sqrt(41))/4 ,x=(-5-sqrt(41))/4
Find the zeros of 2x^2+5x=2
The zeros of 2x^2+5x=2 are x=(-5+sqrt(41))/4 ,x=(-5-sqrt(41))/4