What are the solutions to the equation 5x^2+9x=10 ?

The solutions to the equation 5x^2+9x=10 are x=(-9+sqrt(281))/(10),x=(-9-sqrt(281))/(10)

Find the zeros of 5x^2+9x=10

The zeros of 5x^2+9x=10 are x=(-9+sqrt(281))/(10),x=(-9-sqrt(281))/(10)

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5 x^{2} + 9 x = 10

x = \frac{- 9 + 281}{10}, x = \frac{- 9 - 281}{10}

Decimal Notation

x = 0.77630 \dots, x = - 2.57630 \dots

Solution steps

5 x^{2} + 9 x = 10

Move

10

to the left side

5 x^{2} + 9 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 5 ( - 10 )}{2 \cdot 5}

9^{2} - 4 \cdot 5 (- 10) = 281

x_{1, 2} = \frac{- 9 \pm 281}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 9 + 281}{2 \cdot 5}, x_{2} = \frac{- 9 - 281}{2 \cdot 5}

x = \frac{- 9 + 281}{2 \cdot 5} : \frac{- 9 + 281}{10}

x = \frac{- 9 - 281}{2 \cdot 5} : \frac{- 9 - 281}{10}

The solutions to the quadratic equation are:

x = \frac{- 9 + 281}{10}, x = \frac{- 9 - 281}{10}

a, a^{2} + 6 ab = 16 b^{2}

\frac{3 - x}{1000} = \frac{40 \cdot 10}{2 \cdot 1} (3 - x - 0.7)^{2}

- 5 x^{2} + 40 x - 40 = 0

p^{2} - 3 p - 18 = 0

m^{2} + 4 m + 20 = 0

What are the solutions to the equation 5x^2+9x=10 ?
The solutions to the equation 5x^2+9x=10 are x=(-9+sqrt(281))/(10),x=(-9-sqrt(281))/(10)
Find the zeros of 5x^2+9x=10
The zeros of 5x^2+9x=10 are x=(-9+sqrt(281))/(10),x=(-9-sqrt(281))/(10)