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

The solutions to the equation r^2+r-4=0 are r=(-1+sqrt(17))/2 ,r=(-1-sqrt(17))/2

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

The zeros of r^2+r-4=0 are r=(-1+sqrt(17))/2 ,r=(-1-sqrt(17))/2

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r^{2} + r - 4 = 0

r = \frac{- 1 + 17}{2}, r = \frac{- 1 - 17}{2}

Decimal Notation

r = 1.56155 \dots, r = - 2.56155 \dots

Solution steps

r^{2} + r - 4 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 4) = 17

r_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 1 + 17}{2 \cdot 1}, r_{2} = \frac{- 1 - 17}{2 \cdot 1}

r = \frac{- 1 + 17}{2 \cdot 1} : \frac{- 1 + 17}{2}

r = \frac{- 1 - 17}{2 \cdot 1} : \frac{- 1 - 17}{2}

The solutions to the quadratic equation are:

r = \frac{- 1 + 17}{2}, r = \frac{- 1 - 17}{2}

(h + 1)^{2} = 16

x^{2} + 11 x = 180

a^{2} + 7 a = 0

x^{2} + (\frac{3}{2})^{2} = 9

(x + 9)^{2} + 7 (x + 9) + 6 = 0

What are the solutions to the equation r^2+r-4=0 ?
The solutions to the equation r^2+r-4=0 are r=(-1+sqrt(17))/2 ,r=(-1-sqrt(17))/2
Find the zeros of r^2+r-4=0
The zeros of r^2+r-4=0 are r=(-1+sqrt(17))/2 ,r=(-1-sqrt(17))/2