What are the solutions to the equation 2x^2-5x-9=0 ?

The solutions to the equation 2x^2-5x-9=0 are x=(5+sqrt(97))/4 ,x=(5-sqrt(97))/4

Find the zeros of 2x^2-5x-9=0

The zeros of 2x^2-5x-9=0 are x=(5+sqrt(97))/4 ,x=(5-sqrt(97))/4

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2 x^{2} - 5 x - 9 = 0

x = \frac{5 + 97}{4}, x = \frac{5 - 97}{4}

Decimal Notation

x = 3.71221 \dots, x = - 1.21221 \dots

Solution steps

2 x^{2} - 5 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 2 ( - 9 )}{2 \cdot 2}

(- 5)^{2} - 4 \cdot 2 (- 9) = 97

x_{1, 2} = \frac{- ( - 5 ) \pm 97}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 97}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 97}{2 \cdot 2}

x = \frac{- ( - 5 ) + 97}{2 \cdot 2} : \frac{5 + 97}{4}

x = \frac{- ( - 5 ) - 97}{2 \cdot 2} : \frac{5 - 97}{4}

The solutions to the quadratic equation are:

x = \frac{5 + 97}{4}, x = \frac{5 - 97}{4}

x (x + 4) = 12

20 m^{2} + 13 mn + 2 n^{2}

3 n^{2} - 2 n + 1 = - 3 n^{2} + 9 n + 11

\frac{\frac{25 x ^{2} - y ^{2}}{x y}}{\frac{5}{y} - \frac{1}{x}}

What are the solutions to the equation 2x^2-5x-9=0 ?
The solutions to the equation 2x^2-5x-9=0 are x=(5+sqrt(97))/4 ,x=(5-sqrt(97))/4
Find the zeros of 2x^2-5x-9=0
The zeros of 2x^2-5x-9=0 are x=(5+sqrt(97))/4 ,x=(5-sqrt(97))/4