What are the solutions to the equation 2x^2+3x=7 ?

The solutions to the equation 2x^2+3x=7 are x=(-3+sqrt(65))/4 ,x=(-3-sqrt(65))/4

Find the zeros of 2x^2+3x=7

The zeros of 2x^2+3x=7 are x=(-3+sqrt(65))/4 ,x=(-3-sqrt(65))/4

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2 x^{2} + 3 x = 7

x = \frac{- 3 + 65}{4}, x = \frac{- 3 - 65}{4}

Decimal Notation

x = 1.26556 \dots, x = - 2.76556 \dots

Solution steps

2 x^{2} + 3 x = 7

Move

7

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 7) = 65

x_{1, 2} = \frac{- 3 \pm 65}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 3 + 65}{2 \cdot 2}, x_{2} = \frac{- 3 - 65}{2 \cdot 2}

x = \frac{- 3 + 65}{2 \cdot 2} : \frac{- 3 + 65}{4}

x = \frac{- 3 - 65}{2 \cdot 2} : \frac{- 3 - 65}{4}

The solutions to the quadratic equation are:

x = \frac{- 3 + 65}{4}, x = \frac{- 3 - 65}{4}

2 z^{2} + 3 z - 20

- 4 \leq x + 5 \leq 5

ln (\frac{1}{e ^{5}}) = y

(x + 2) (x - 3) = 0

10 (1 + 3 b) = - 20

What are the solutions to the equation 2x^2+3x=7 ?
The solutions to the equation 2x^2+3x=7 are x=(-3+sqrt(65))/4 ,x=(-3-sqrt(65))/4
Find the zeros of 2x^2+3x=7
The zeros of 2x^2+3x=7 are x=(-3+sqrt(65))/4 ,x=(-3-sqrt(65))/4