What are the solutions to the equation 4x^2-19x-5=0 ?

The solutions to the equation 4x^2-19x-5=0 are x=5,x=-1/4

Find the zeros of 4x^2-19x-5=0

The zeros of 4x^2-19x-5=0 are x=5,x=-1/4

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4 x^{2} - 19 x - 5 = 0

x = 5, x = - \frac{1}{4}

Solution steps

4 x^{2} - 19 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

(- 19)^{2} - 4 \cdot 4 (- 5) = 21

x_{1, 2} = \frac{- ( - 19 ) \pm 21}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 19 ) + 21}{2 \cdot 4}, x_{2} = \frac{- ( - 19 ) - 21}{2 \cdot 4}

x = \frac{- ( - 19 ) + 21}{2 \cdot 4} : 5

x = \frac{- ( - 19 ) - 21}{2 \cdot 4} : - \frac{1}{4}

The solutions to the quadratic equation are:

x = 5, x = - \frac{1}{4}

2 x + 5 < 15

\frac{( 3 x + 2 )}{x - 1} > x

x (3 - 4 x) \leq 0

x^{3} + x^{2} - 11 x + 10

7 + s = 8 s - (- 9)

What are the solutions to the equation 4x^2-19x-5=0 ?
The solutions to the equation 4x^2-19x-5=0 are x=5,x=-1/4
Find the zeros of 4x^2-19x-5=0
The zeros of 4x^2-19x-5=0 are x=5,x=-1/4