What are the solutions to the equation 4x^2+17=-20x ?

The solutions to the equation 4x^2+17=-20x are x=(-5+2sqrt(2))/2 ,x=-(5+2sqrt(2))/2

Find the zeros of 4x^2+17=-20x

The zeros of 4x^2+17=-20x are x=(-5+2sqrt(2))/2 ,x=-(5+2sqrt(2))/2

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4 x^{2} + 17 = - 20 x

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

Solution steps

4 x^{2} + 17 = - 20 x

Move

20 x

to the left side

4 x^{2} + 17 + 20 x = 0

Write in the standard form

a x^{2} + b x + c = 0

4 x^{2} + 20 x + 17 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 17}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 \cdot 17 = 82

x_{1, 2} = \frac{- 20 \pm 8 2}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 8 2}{2 \cdot 4}, x_{2} = \frac{- 20 - 8 2}{2 \cdot 4}

x = \frac{- 20 + 8 2}{2 \cdot 4} : \frac{- 5 + 2 2}{2}

x = \frac{- 20 - 8 2}{2 \cdot 4} : - \frac{5 + 2 2}{2}

The solutions to the quadratic equation are:

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

\frac{2}{x - 1} > \frac{7}{x + 1}

\frac{11 !}{( 11 - 5 ) ! 5 !}

3 x^{2} - 14 x - 49

4 (x + 3) \geq 0 or 2 (x + 4) \leq 4

8 - \frac{9 x}{2} < 5

What are the solutions to the equation 4x^2+17=-20x ?
The solutions to the equation 4x^2+17=-20x are x=(-5+2sqrt(2))/2 ,x=-(5+2sqrt(2))/2
Find the zeros of 4x^2+17=-20x
The zeros of 4x^2+17=-20x are x=(-5+2sqrt(2))/2 ,x=-(5+2sqrt(2))/2