What are the solutions to the equation -32/400 x^2+x+4=0 ?

The solutions to the equation -32/400 x^2+x+4=0 are x=-(5(sqrt(57)-5))/4 ,x=(5(5+sqrt(57)))/4

Find the zeros of -32/400 x^2+x+4=0

The zeros of -32/400 x^2+x+4=0 are x=-(5(sqrt(57)-5))/4 ,x=(5(5+sqrt(57)))/4

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- \frac{32}{400} x^{2} + x + 4 = 0

x = - \frac{5 ( 57 - 5 )}{4}, x = \frac{5 ( 5 + 57 )}{4}

Decimal Notation

x = - 3.18729 \dots, x = 15.68729 \dots

Solution steps

- \frac{32}{400} x^{2} + x + 4 = 0

Multiply both sides by

400

- 32 x^{2} + 400 x + 1600 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 400 \pm 40 0 ^{2} - 4 ( - 32 ) \cdot 1600}{2 ( - 32 )}

40 0^{2} - 4 (- 32) \cdot 1600 = 8057

x_{1, 2} = \frac{- 400 \pm 80 57}{2 ( - 32 )}

Separate the solutions

x_{1} = \frac{- 400 + 80 57}{2 ( - 32 )}, x_{2} = \frac{- 400 - 80 57}{2 ( - 32 )}

x = \frac{- 400 + 80 57}{2 ( - 32 )} : - \frac{5 ( 57 - 5 )}{4}

x = \frac{- 400 - 80 57}{2 ( - 32 )} : \frac{5 ( 5 + 57 )}{4}

The solutions to the quadratic equation are:

x = - \frac{5 ( 57 - 5 )}{4}, x = \frac{5 ( 5 + 57 )}{4}

2 x^{2} + 3 x = 2

\frac{n}{2} < 0

10 w (w - 2) + 10 = 0

x - 4 \geq 17

1 + \frac{2 x}{7} - \frac{1}{2} (x + 7) \geq x

What are the solutions to the equation -32/400 x^2+x+4=0 ?
The solutions to the equation -32/400 x^2+x+4=0 are x=-(5(sqrt(57)-5))/4 ,x=(5(5+sqrt(57)))/4
Find the zeros of -32/400 x^2+x+4=0
The zeros of -32/400 x^2+x+4=0 are x=-(5(sqrt(57)-5))/4 ,x=(5(5+sqrt(57)))/4