What are the solutions to the equation 2r^2+r-14=0 ?

The solutions to the equation 2r^2+r-14=0 are r=(-1+sqrt(113))/4 ,r=(-1-sqrt(113))/4

Find the zeros of 2r^2+r-14=0

The zeros of 2r^2+r-14=0 are r=(-1+sqrt(113))/4 ,r=(-1-sqrt(113))/4

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2 r^{2} + r - 14 = 0

r = \frac{- 1 + 113}{4}, r = \frac{- 1 - 113}{4}

Decimal Notation

r = 2.40753 \dots, r = - 2.90753 \dots

Solution steps

2 r^{2} + r - 14 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 14 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 14) = 113

r_{1, 2} = \frac{- 1 \pm 113}{2 \cdot 2}

Separate the solutions

r_{1} = \frac{- 1 + 113}{2 \cdot 2}, r_{2} = \frac{- 1 - 113}{2 \cdot 2}

r = \frac{- 1 + 113}{2 \cdot 2} : \frac{- 1 + 113}{4}

r = \frac{- 1 - 113}{2 \cdot 2} : \frac{- 1 - 113}{4}

The solutions to the quadratic equation are:

r = \frac{- 1 + 113}{4}, r = \frac{- 1 - 113}{4}

\frac{3 x + 1}{3} - \frac{5 x ^{2} + 3}{2} = \frac{x ^{2} - 1}{2} - \frac{x + 2}{3}

x, x^{2} + y^{2} = 4 x y

3 x^{2} - 24 x + 40 = 0

x^{2} - 10 = 2 x^{2}

n^{2} + 6 n + 9 = 0

What are the solutions to the equation 2r^2+r-14=0 ?
The solutions to the equation 2r^2+r-14=0 are r=(-1+sqrt(113))/4 ,r=(-1-sqrt(113))/4
Find the zeros of 2r^2+r-14=0
The zeros of 2r^2+r-14=0 are r=(-1+sqrt(113))/4 ,r=(-1-sqrt(113))/4