What are the solutions to the equation 2x^2-15x-2=0 ?

The solutions to the equation 2x^2-15x-2=0 are x=(15+sqrt(241))/4 ,x=(15-sqrt(241))/4

Find the zeros of 2x^2-15x-2=0

The zeros of 2x^2-15x-2=0 are x=(15+sqrt(241))/4 ,x=(15-sqrt(241))/4

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

x = \frac{15 + 241}{4}, x = \frac{15 - 241}{4}

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

(- 15)^{2} - 4 \cdot 2 (- 2) = 241

x_{1, 2} = \frac{- ( - 15 ) \pm 241}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 15 ) + 241}{2 \cdot 2}, x_{2} = \frac{- ( - 15 ) - 241}{2 \cdot 2}

x = \frac{- ( - 15 ) + 241}{2 \cdot 2} : \frac{15 + 241}{4}

x = \frac{- ( - 15 ) - 241}{2 \cdot 2} : \frac{15 - 241}{4}

The solutions to the quadratic equation are:

x = \frac{15 + 241}{4}, x = \frac{15 - 241}{4}

x^{4} + x^{3} - 8 x^{2} + 2 x + 4

- 2 x^{2} + 4 x + 30

\frac{2 x + 1}{3 x - 2} \geq 0

(x - 1)^{2} \geq 0

6 (\frac{11 - 4 y}{2}) - 5 = 10 - 6 y

What are the solutions to the equation 2x^2-15x-2=0 ?
The solutions to the equation 2x^2-15x-2=0 are x=(15+sqrt(241))/4 ,x=(15-sqrt(241))/4
Find the zeros of 2x^2-15x-2=0
The zeros of 2x^2-15x-2=0 are x=(15+sqrt(241))/4 ,x=(15-sqrt(241))/4