What are the solutions to the equation 3=5(x^2+2x) ?

The solutions to the equation 3=5(x^2+2x) are x=(-5+2sqrt(10))/5 ,x=-(5+2sqrt(10))/5

Find the zeros of 3=5(x^2+2x)

The zeros of 3=5(x^2+2x) are x=(-5+2sqrt(10))/5 ,x=-(5+2sqrt(10))/5

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3 = 5 (x^{2} + 2 x)

x = \frac{- 5 + 2 10}{5}, x = - \frac{5 + 2 10}{5}

Decimal Notation

x = 0.26491 \dots, x = - 2.26491 \dots

Solution steps

3 = 5 (x^{2} + 2 x)

Expand

5 (x^{2} + 2 x) : 5 x^{2} + 10 x

3 = 5 x^{2} + 10 x

Switch sides

5 x^{2} + 10 x = 3

Move

3

to the left side

5 x^{2} + 10 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

1 0^{2} - 4 \cdot 5 (- 3) = 410

x_{1, 2} = \frac{- 10 \pm 4 10}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 10 + 4 10}{2 \cdot 5}, x_{2} = \frac{- 10 - 4 10}{2 \cdot 5}

x = \frac{- 10 + 4 10}{2 \cdot 5} : \frac{- 5 + 2 10}{5}

x = \frac{- 10 - 4 10}{2 \cdot 5} : - \frac{5 + 2 10}{5}

The solutions to the quadratic equation are:

x = \frac{- 5 + 2 10}{5}, x = - \frac{5 + 2 10}{5}

F = m v^{2}

0 = - 6 x^{2} + 54

- r^{2} - 18 = 3 r - 2 r^{2}

x^{2} - 2 x + 49 = 11 x + 7

6 x^{2} - 12 x + 7 = 0

What are the solutions to the equation 3=5(x^2+2x) ?
The solutions to the equation 3=5(x^2+2x) are x=(-5+2sqrt(10))/5 ,x=-(5+2sqrt(10))/5
Find the zeros of 3=5(x^2+2x)
The zeros of 3=5(x^2+2x) are x=(-5+2sqrt(10))/5 ,x=-(5+2sqrt(10))/5