What are the solutions to the equation 2x^2+7x=5 ?

The solutions to the equation 2x^2+7x=5 are x=(-7+sqrt(89))/4 ,x=(-7-sqrt(89))/4

Find the zeros of 2x^2+7x=5

The zeros of 2x^2+7x=5 are x=(-7+sqrt(89))/4 ,x=(-7-sqrt(89))/4

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2 x^{2} + 7 x = 5

x = \frac{- 7 + 89}{4}, x = \frac{- 7 - 89}{4}

Decimal Notation

x = 0.60849 \dots, x = - 4.10849 \dots

Solution steps

2 x^{2} + 7 x = 5

Move

5

to the left side

2 x^{2} + 7 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 ( - 5 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 5) = 89

x_{1, 2} = \frac{- 7 \pm 89}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 89}{2 \cdot 2}, x_{2} = \frac{- 7 - 89}{2 \cdot 2}

x = \frac{- 7 + 89}{2 \cdot 2} : \frac{- 7 + 89}{4}

x = \frac{- 7 - 89}{2 \cdot 2} : \frac{- 7 - 89}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 89}{4}, x = \frac{- 7 - 89}{4}

13384.63 (\frac{0.075}{12})

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

- 7 x + 1 \geq 22

\frac{( 2 + i )}{i}

729 y^{3} - 1

What are the solutions to the equation 2x^2+7x=5 ?
The solutions to the equation 2x^2+7x=5 are x=(-7+sqrt(89))/4 ,x=(-7-sqrt(89))/4
Find the zeros of 2x^2+7x=5
The zeros of 2x^2+7x=5 are x=(-7+sqrt(89))/4 ,x=(-7-sqrt(89))/4