What are the solutions to the equation 2x^2+19x-10=0 ?

The solutions to the equation 2x^2+19x-10=0 are x= 1/2 ,x=-10

Find the zeros of 2x^2+19x-10=0

The zeros of 2x^2+19x-10=0 are x= 1/2 ,x=-10

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

x = \frac{1}{2}, x = - 10

Decimal Notation

x = 0.5, x = - 10

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 2 ( - 10 )}{2 \cdot 2}

1 9^{2} - 4 \cdot 2 (- 10) = 21

x_{1, 2} = \frac{- 19 \pm 21}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 19 + 21}{2 \cdot 2}, x_{2} = \frac{- 19 - 21}{2 \cdot 2}

x = \frac{- 19 + 21}{2 \cdot 2} : \frac{1}{2}

x = \frac{- 19 - 21}{2 \cdot 2} : - 10

The solutions to the quadratic equation are:

x = \frac{1}{2}, x = - 10

\frac{1}{2} x + 4 = \frac{3}{2} x + 5

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

(- \frac{3}{2})^{2} + 6 (- \frac{3}{2})

\frac{( 49 - 33.3 )}{11.2}

14 + 13 y = 20 y - 21

What are the solutions to the equation 2x^2+19x-10=0 ?
The solutions to the equation 2x^2+19x-10=0 are x= 1/2 ,x=-10
Find the zeros of 2x^2+19x-10=0
The zeros of 2x^2+19x-10=0 are x= 1/2 ,x=-10