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

The solutions to the equation 2x^2-5x-5=0 are x=(5+sqrt(65))/4 ,x=(5-sqrt(65))/4

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

The zeros of 2x^2-5x-5=0 are x=(5+sqrt(65))/4 ,x=(5-sqrt(65))/4

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

x = \frac{5 + 65}{4}, x = \frac{5 - 65}{4}

Decimal Notation

x = 3.26556 \dots, x = - 0.76556 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 5)^{2} - 4 \cdot 2 (- 5) = 65

x_{1, 2} = \frac{- ( - 5 ) \pm 65}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 65}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 65}{2 \cdot 2}

x = \frac{- ( - 5 ) + 65}{2 \cdot 2} : \frac{5 + 65}{4}

x = \frac{- ( - 5 ) - 65}{2 \cdot 2} : \frac{5 - 65}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 65}{4}, x = \frac{5 - 65}{4}

- 1

\frac{( 3 x ^{5} y ^{5} ) ^{3}}{3 x ^{3} y ^{11}} + 4 x^{12} y^{4}

x^{2} y^{4} - 49

4 \cdot 6

x^{2} - x - 182 = 0

What are the solutions to the equation 2x^2-5x-5=0 ?
The solutions to the equation 2x^2-5x-5=0 are x=(5+sqrt(65))/4 ,x=(5-sqrt(65))/4
Find the zeros of 2x^2-5x-5=0
The zeros of 2x^2-5x-5=0 are x=(5+sqrt(65))/4 ,x=(5-sqrt(65))/4