What are the solutions to the equation -10x^2-2x+3=0 ?

The solutions to the equation -10x^2-2x+3=0 are x=-(1+sqrt(31))/(10),x=(sqrt(31)-1)/(10)

Find the zeros of -10x^2-2x+3=0

The zeros of -10x^2-2x+3=0 are x=-(1+sqrt(31))/(10),x=(sqrt(31)-1)/(10)

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

x = - \frac{1 + 31}{10}, x = \frac{31 - 1}{10}

Decimal Notation

x = - 0.65677 \dots, x = 0.45677 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 (- 10) \cdot 3 = 231

x_{1, 2} = \frac{- ( - 2 ) \pm 2 31}{2 ( - 10 )}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 31}{2 ( - 10 )}, x_{2} = \frac{- ( - 2 ) - 2 31}{2 ( - 10 )}

x = \frac{- ( - 2 ) + 2 31}{2 ( - 10 )} : - \frac{1 + 31}{10}

x = \frac{- ( - 2 ) - 2 31}{2 ( - 10 )} : \frac{31 - 1}{10}

The solutions to the quadratic equation are:

x = - \frac{1 + 31}{10}, x = \frac{31 - 1}{10}

p^{2} - 1 = 11

\frac{( x + 2 )}{( 4 )} = \frac{( ( x + 1 ) ( x + 2 ) )}{( 2 )}

- 0.1 q^{2} + 90 = 0.2 q^{2} + q + 50

(3 x - 5)^{2} = 0

2 x^{2} - 13 x - 6 = 0

What are the solutions to the equation -10x^2-2x+3=0 ?
The solutions to the equation -10x^2-2x+3=0 are x=-(1+sqrt(31))/(10),x=(sqrt(31)-1)/(10)
Find the zeros of -10x^2-2x+3=0
The zeros of -10x^2-2x+3=0 are x=-(1+sqrt(31))/(10),x=(sqrt(31)-1)/(10)