What are the solutions to the equation -5x^2+2x+11=0 ?

The solutions to the equation -5x^2+2x+11=0 are x=-(-1+2sqrt(14))/5 ,x=(1+2sqrt(14))/5

Find the zeros of -5x^2+2x+11=0

The zeros of -5x^2+2x+11=0 are x=-(-1+2sqrt(14))/5 ,x=(1+2sqrt(14))/5

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- 5 x^{2} + 2 x + 11 = 0

x = - \frac{- 1 + 2 14}{5}, x = \frac{1 + 2 14}{5}

Decimal Notation

x = - 1.29666 \dots, x = 1.69666 \dots

Solution steps

- 5 x^{2} + 2 x + 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 5 ) \cdot 11}{2 ( - 5 )}

2^{2} - 4 (- 5) \cdot 11 = 414

x_{1, 2} = \frac{- 2 \pm 4 14}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- 2 + 4 14}{2 ( - 5 )}, x_{2} = \frac{- 2 - 4 14}{2 ( - 5 )}

x = \frac{- 2 + 4 14}{2 ( - 5 )} : - \frac{- 1 + 2 14}{5}

x = \frac{- 2 - 4 14}{2 ( - 5 )} : \frac{1 + 2 14}{5}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 2 14}{5}, x = \frac{1 + 2 14}{5}

2 x^{2} - 2 x + 15 = 0

837.2 = 0.68 \cdot 2.14 \cdot 100 \cdot x \cdot (9.5 - 0.4 \cdot x)

12 x^{2} - 48 x = - 39

(t + 1)^{2} = - 9

y^{2} - 5 y

What are the solutions to the equation -5x^2+2x+11=0 ?
The solutions to the equation -5x^2+2x+11=0 are x=-(-1+2sqrt(14))/5 ,x=(1+2sqrt(14))/5
Find the zeros of -5x^2+2x+11=0
The zeros of -5x^2+2x+11=0 are x=-(-1+2sqrt(14))/5 ,x=(1+2sqrt(14))/5