What are the solutions to the equation r^2-4r-91+7=0 ?

The solutions to the equation r^2-4r-91+7=0 are r=2(1+sqrt(22)),r=2(1-sqrt(22))

Find the zeros of r^2-4r-91+7=0

The zeros of r^2-4r-91+7=0 are r=2(1+sqrt(22)),r=2(1-sqrt(22))

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r^{2} - 4 r - 91 + 7 = 0

r = 2 (1 + 22), r = 2 (1 - 22)

Decimal Notation

r = 11.38083 \dots, r = - 7.38083 \dots

Solution steps

r^{2} - 4 r - 91 + 7 = 0

Refine

r^{2} - 4 r - 84 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 84 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 84) = 422

r_{1, 2} = \frac{- ( - 4 ) \pm 4 22}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 4 ) + 4 22}{2 \cdot 1}, r_{2} = \frac{- ( - 4 ) - 4 22}{2 \cdot 1}

r = \frac{- ( - 4 ) + 4 22}{2 \cdot 1} : 2 (1 + 22)

r = \frac{- ( - 4 ) - 4 22}{2 \cdot 1} : 2 (1 - 22)

The solutions to the quadratic equation are:

r = 2 (1 + 22), r = 2 (1 - 22)

3 x^{2} - 5 x - 16 = 0

4 x^{2} - 28 x + 49

m^{2} - 5 m - 14 = 0

5 n^{2} - 14 = - 6

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

What are the solutions to the equation r^2-4r-91+7=0 ?
The solutions to the equation r^2-4r-91+7=0 are r=2(1+sqrt(22)),r=2(1-sqrt(22))
Find the zeros of r^2-4r-91+7=0
The zeros of r^2-4r-91+7=0 are r=2(1+sqrt(22)),r=2(1-sqrt(22))