What are the solutions to the equation -9x+10x^2+8=14 ?

The solutions to the equation -9x+10x^2+8=14 are x=(9+sqrt(321))/(20),x=(9-sqrt(321))/(20)

Find the zeros of -9x+10x^2+8=14

The zeros of -9x+10x^2+8=14 are x=(9+sqrt(321))/(20),x=(9-sqrt(321))/(20)

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- 9 x + 10 x^{2} + 8 = 14

x = \frac{9 + 321}{20}, x = \frac{9 - 321}{20}

Decimal Notation

x = 1.34582 \dots, x = - 0.44582 \dots

Solution steps

- 9 x + 10 x^{2} + 8 = 14

Move

14

to the left side

10 x^{2} - 9 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 10 ( - 6 )}{2 \cdot 10}

(- 9)^{2} - 4 \cdot 10 (- 6) = 321

x_{1, 2} = \frac{- ( - 9 ) \pm 321}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 321}{2 \cdot 10}, x_{2} = \frac{- ( - 9 ) - 321}{2 \cdot 10}

x = \frac{- ( - 9 ) + 321}{2 \cdot 10} : \frac{9 + 321}{20}

x = \frac{- ( - 9 ) - 321}{2 \cdot 10} : \frac{9 - 321}{20}

The solutions to the quadratic equation are:

x = \frac{9 + 321}{20}, x = \frac{9 - 321}{20}

\frac{( 8 n - 1 ) !}{( 8 n + 1 ) !}

7 (5 a - 4) - 1 = 14 - 8 a

2 (x - 5) = 3 x + 3

\frac{2}{9} (x - 1) \leq \frac{x}{18} + \frac{2}{9}

y (ab + 4) - 3 (4 + ab)

What are the solutions to the equation -9x+10x^2+8=14 ?
The solutions to the equation -9x+10x^2+8=14 are x=(9+sqrt(321))/(20),x=(9-sqrt(321))/(20)
Find the zeros of -9x+10x^2+8=14
The zeros of -9x+10x^2+8=14 are x=(9+sqrt(321))/(20),x=(9-sqrt(321))/(20)