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

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

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

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

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

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

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

1 9^{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} : \frac{1}{4}

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

The solutions to the quadratic equation are:

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

0 - 6

\frac{a ^{3} b ^{- 2}}{a b ^{- 4}}

- \frac{5}{8} \cdot 3

10 = t - 8

(\frac{c ^{4} d ^{- 3}}{c ^{7} d ^{5}})^{\frac{1}{4}}

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