What are the solutions to the equation 2x^2-5x-7=0 ?

The solutions to the equation 2x^2-5x-7=0 are x= 7/2 ,x=-1

Find the zeros of 2x^2-5x-7=0

The zeros of 2x^2-5x-7=0 are x= 7/2 ,x=-1

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2 x^{2} - 5 x - 7 = 0

x = \frac{7}{2}, x = - 1

Decimal Notation

x = 3.5, x = - 1

Solution steps

2 x^{2} - 5 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 (- 7) = 9

x_{1, 2} = \frac{- ( - 5 ) \pm 9}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 9}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 9}{2 \cdot 2}

x = \frac{- ( - 5 ) + 9}{2 \cdot 2} : \frac{7}{2}

x = \frac{- ( - 5 ) - 9}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

x = \frac{7}{2}, x = - 1

36 \cdot \frac{π}{180}

(2 x^{2})^{3}

3.6 - 4.85 a \geq - 6.1

2 x - 6 = 2

5 a \leq 10

What are the solutions to the equation 2x^2-5x-7=0 ?
The solutions to the equation 2x^2-5x-7=0 are x= 7/2 ,x=-1
Find the zeros of 2x^2-5x-7=0
The zeros of 2x^2-5x-7=0 are x= 7/2 ,x=-1