What are the solutions to the equation 5x^2-7x-4=0 ?

The solutions to the equation 5x^2-7x-4=0 are x=(7+sqrt(129))/(10),x=(7-sqrt(129))/(10)

Find the zeros of 5x^2-7x-4=0

The zeros of 5x^2-7x-4=0 are x=(7+sqrt(129))/(10),x=(7-sqrt(129))/(10)

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5 x^{2} - 7 x - 4 = 0

x = \frac{7 + 129}{10}, x = \frac{7 - 129}{10}

Decimal Notation

x = 1.83578 \dots, x = - 0.43578 \dots

Solution steps

5 x^{2} - 7 x - 4 = 0

Solve with the quadratic formula

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

(- 7)^{2} - 4 \cdot 5 (- 4) = 129

x_{1, 2} = \frac{- ( - 7 ) \pm 129}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 129}{2 \cdot 5}, x_{2} = \frac{- ( - 7 ) - 129}{2 \cdot 5}

x = \frac{- ( - 7 ) + 129}{2 \cdot 5} : \frac{7 + 129}{10}

x = \frac{- ( - 7 ) - 129}{2 \cdot 5} : \frac{7 - 129}{10}

The solutions to the quadratic equation are:

x = \frac{7 + 129}{10}, x = \frac{7 - 129}{10}

((1 + x)^{\frac{1}{2}})^{2}

4^{2} + b^{2} = 5^{2}

\frac{2 x y}{x ^{2} - y ^{2}}

- 5 \leq - 10 x < 5

What are the solutions to the equation 5x^2-7x-4=0 ?
The solutions to the equation 5x^2-7x-4=0 are x=(7+sqrt(129))/(10),x=(7-sqrt(129))/(10)
Find the zeros of 5x^2-7x-4=0
The zeros of 5x^2-7x-4=0 are x=(7+sqrt(129))/(10),x=(7-sqrt(129))/(10)