What are the solutions to the equation 5x^2-3x-5=0 ?

The solutions to the equation 5x^2-3x-5=0 are x=(3+sqrt(109))/(10),x=(3-sqrt(109))/(10)

Find the zeros of 5x^2-3x-5=0

The zeros of 5x^2-3x-5=0 are x=(3+sqrt(109))/(10),x=(3-sqrt(109))/(10)

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5 x^{2} - 3 x - 5 = 0

x = \frac{3 + 109}{10}, x = \frac{3 - 109}{10}

Decimal Notation

x = 1.34403 \dots, x = - 0.74403 \dots

Solution steps

5 x^{2} - 3 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

(- 3)^{2} - 4 \cdot 5 (- 5) = 109

x_{1, 2} = \frac{- ( - 3 ) \pm 109}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 109}{2 \cdot 5}, x_{2} = \frac{- ( - 3 ) - 109}{2 \cdot 5}

x = \frac{- ( - 3 ) + 109}{2 \cdot 5} : \frac{3 + 109}{10}

x = \frac{- ( - 3 ) - 109}{2 \cdot 5} : \frac{3 - 109}{10}

The solutions to the quadratic equation are:

x = \frac{3 + 109}{10}, x = \frac{3 - 109}{10}

\frac{y}{4} - \frac{y + 1}{5} = \frac{y + 2}{8}

16 x^{3} - 9 x^{2} - 7 x

\frac{4 x}{1 - x ^{2}} \leq \frac{3 x}{1 - x}

e^{- 0.03}

What are the solutions to the equation 5x^2-3x-5=0 ?
The solutions to the equation 5x^2-3x-5=0 are x=(3+sqrt(109))/(10),x=(3-sqrt(109))/(10)
Find the zeros of 5x^2-3x-5=0
The zeros of 5x^2-3x-5=0 are x=(3+sqrt(109))/(10),x=(3-sqrt(109))/(10)