What are the solutions to the equation 28x^2-5x-4=0 ?

The solutions to the equation 28x^2-5x-4=0 are x=(5+sqrt(473))/(56),x=(5-sqrt(473))/(56)

Find the zeros of 28x^2-5x-4=0

The zeros of 28x^2-5x-4=0 are x=(5+sqrt(473))/(56),x=(5-sqrt(473))/(56)

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

x = \frac{5 + 473}{56}, x = \frac{5 - 473}{56}

Decimal Notation

x = 0.47765 \dots, x = - 0.29908 \dots

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 28 ( - 4 )}{2 \cdot 28}

(- 5)^{2} - 4 \cdot 28 (- 4) = 473

x_{1, 2} = \frac{- ( - 5 ) \pm 473}{2 \cdot 28}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 473}{2 \cdot 28}, x_{2} = \frac{- ( - 5 ) - 473}{2 \cdot 28}

x = \frac{- ( - 5 ) + 473}{2 \cdot 28} : \frac{5 + 473}{56}

x = \frac{- ( - 5 ) - 473}{2 \cdot 28} : \frac{5 - 473}{56}

The solutions to the quadratic equation are:

x = \frac{5 + 473}{56}, x = \frac{5 - 473}{56}

(x + 2)^{2} - 36

\frac{1}{2} (2 h - 4) = 20

x^{3} + 6 x^{2} - 9 x - 54

(- 5 x)^{2}

(3)^{10} (3)^{- 9}

What are the solutions to the equation 28x^2-5x-4=0 ?
The solutions to the equation 28x^2-5x-4=0 are x=(5+sqrt(473))/(56),x=(5-sqrt(473))/(56)
Find the zeros of 28x^2-5x-4=0
The zeros of 28x^2-5x-4=0 are x=(5+sqrt(473))/(56),x=(5-sqrt(473))/(56)