What are the solutions to the equation 10x^2-20x=19 ?

The solutions to the equation 10x^2-20x=19 are x=(10+sqrt(290))/(10),x=(10-sqrt(290))/(10)

Find the zeros of 10x^2-20x=19

The zeros of 10x^2-20x=19 are x=(10+sqrt(290))/(10),x=(10-sqrt(290))/(10)

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10 x^{2} - 20 x = 19

x = \frac{10 + 290}{10}, x = \frac{10 - 290}{10}

Decimal Notation

x = 2.70293 \dots, x = - 0.70293 \dots

Solution steps

10 x^{2} - 20 x = 19

Move

19

to the left side

10 x^{2} - 20 x - 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 10 ( - 19 )}{2 \cdot 10}

(- 20)^{2} - 4 \cdot 10 (- 19) = 2290

x_{1, 2} = \frac{- ( - 20 ) \pm 2 290}{2 \cdot 10}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 2 290}{2 \cdot 10}, x_{2} = \frac{- ( - 20 ) - 2 290}{2 \cdot 10}

x = \frac{- ( - 20 ) + 2 290}{2 \cdot 10} : \frac{10 + 290}{10}

x = \frac{- ( - 20 ) - 2 290}{2 \cdot 10} : \frac{10 - 290}{10}

The solutions to the quadratic equation are:

x = \frac{10 + 290}{10}, x = \frac{10 - 290}{10}

- x^{2} + 6 x + 18 = 0

t, d = \frac{1}{2} g t^{2}

- 5 x^{2} - 80 x - 140 = 0

6 x^{2} - 4 x = 2

6 m^{2} + 26 m = 20

What are the solutions to the equation 10x^2-20x=19 ?
The solutions to the equation 10x^2-20x=19 are x=(10+sqrt(290))/(10),x=(10-sqrt(290))/(10)
Find the zeros of 10x^2-20x=19
The zeros of 10x^2-20x=19 are x=(10+sqrt(290))/(10),x=(10-sqrt(290))/(10)