What are the solutions to the equation 2r^2+8r-19=0 ?

The solutions to the equation 2r^2+8r-19=0 are r=(-4+3sqrt(6))/2 ,r=-(4+3sqrt(6))/2

Find the zeros of 2r^2+8r-19=0

The zeros of 2r^2+8r-19=0 are r=(-4+3sqrt(6))/2 ,r=-(4+3sqrt(6))/2

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2 r^{2} + 8 r - 19 = 0

r = \frac{- 4 + 3 6}{2}, r = - \frac{4 + 3 6}{2}

Decimal Notation

r = 1.67423 \dots, r = - 5.67423 \dots

Solution steps

2 r^{2} + 8 r - 19 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 ( - 19 )}{2 \cdot 2}

8^{2} - 4 \cdot 2 (- 19) = 66

r_{1, 2} = \frac{- 8 \pm 6 6}{2 \cdot 2}

Separate the solutions

r_{1} = \frac{- 8 + 6 6}{2 \cdot 2}, r_{2} = \frac{- 8 - 6 6}{2 \cdot 2}

r = \frac{- 8 + 6 6}{2 \cdot 2} : \frac{- 4 + 3 6}{2}

r = \frac{- 8 - 6 6}{2 \cdot 2} : - \frac{4 + 3 6}{2}

The solutions to the quadratic equation are:

r = \frac{- 4 + 3 6}{2}, r = - \frac{4 + 3 6}{2}

(- \frac{3}{2})^{2} + (\frac{1}{2})^{2} = r^{2}

3 x^{2} - 6 x + 2 = 0

2 x^{2} - 8 = 41

3 x^{2} - x - 48 = 0

- 0.6 t^{2} + 48 t = 0

What are the solutions to the equation 2r^2+8r-19=0 ?
The solutions to the equation 2r^2+8r-19=0 are r=(-4+3sqrt(6))/2 ,r=-(4+3sqrt(6))/2
Find the zeros of 2r^2+8r-19=0
The zeros of 2r^2+8r-19=0 are r=(-4+3sqrt(6))/2 ,r=-(4+3sqrt(6))/2