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

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

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

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

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

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

Decimal Notation

r = 0.25, r = - 5

Solution steps

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

Solve with the quadratic formula

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

1 9^{2} - 4 \cdot 4 (- 5) = 21

r_{1, 2} = \frac{- 19 \pm 21}{2 \cdot 4}

Separate the solutions

r_{1} = \frac{- 19 + 21}{2 \cdot 4}, r_{2} = \frac{- 19 - 21}{2 \cdot 4}

r = \frac{- 19 + 21}{2 \cdot 4} : \frac{1}{4}

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

The solutions to the quadratic equation are:

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

2 1^{2} + 2 8^{2}

- 7 - 10175

2 x + 3 = 6 x - 4

53 t - 5 t^{2} - 2 = 0

\frac{1}{4} + \frac{2}{5}

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