What are the solutions to the equation 2x^2+5x=4 ?

The solutions to the equation 2x^2+5x=4 are x=(-5+sqrt(57))/4 ,x=(-5-sqrt(57))/4

Find the zeros of 2x^2+5x=4

The zeros of 2x^2+5x=4 are x=(-5+sqrt(57))/4 ,x=(-5-sqrt(57))/4

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2 x^{2} + 5 x = 4

x = \frac{- 5 + 57}{4}, x = \frac{- 5 - 57}{4}

Decimal Notation

x = 0.63745 \dots, x = - 3.13745 \dots

Solution steps

2 x^{2} + 5 x = 4

Move

4

to the left side

2 x^{2} + 5 x - 4 = 0

Solve with the quadratic formula

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

5^{2} - 4 \cdot 2 (- 4) = 57

x_{1, 2} = \frac{- 5 \pm 57}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 57}{2 \cdot 2}, x_{2} = \frac{- 5 - 57}{2 \cdot 2}

x = \frac{- 5 + 57}{2 \cdot 2} : \frac{- 5 + 57}{4}

x = \frac{- 5 - 57}{2 \cdot 2} : \frac{- 5 - 57}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 57}{4}, x = \frac{- 5 - 57}{4}

12 x^{2} - 7 x - 12

350 - 272 + 24

(\frac{7}{9})^{- 2}

3 \cdot \frac{5}{6}

- 1 \cdot \frac{1}{2}

What are the solutions to the equation 2x^2+5x=4 ?
The solutions to the equation 2x^2+5x=4 are x=(-5+sqrt(57))/4 ,x=(-5-sqrt(57))/4
Find the zeros of 2x^2+5x=4
The zeros of 2x^2+5x=4 are x=(-5+sqrt(57))/4 ,x=(-5-sqrt(57))/4