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

The solutions to the equation r^2-4r-91=7 are r=2+sqrt(102),r=2-sqrt(102)

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

The zeros of r^2-4r-91=7 are r=2+sqrt(102),r=2-sqrt(102)

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

r = 2 + 102, r = 2 - 102

Solution steps

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

Move

7

to the left side

r^{2} - 4 r - 98 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 98 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 98) = 2102

r_{1, 2} = \frac{- ( - 4 ) \pm 2 102}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 4 ) + 2 102}{2 \cdot 1}, r_{2} = \frac{- ( - 4 ) - 2 102}{2 \cdot 1}

r = \frac{- ( - 4 ) + 2 102}{2 \cdot 1} : 2 + 102

r = \frac{- ( - 4 ) - 2 102}{2 \cdot 1} : 2 - 102

The solutions to the quadratic equation are:

r = 2 + 102, r = 2 - 102

5 + \frac{1}{3}

3 y^{2} + 5 y = 12

1 2^{\frac{1}{8}} \cdot 1 2^{\frac{5}{6}}

(x + 6) (4 x - 7) = 0

∣ x - 9 ∣ = - 3

What are the solutions to the equation r^2-4r-91=7 ?
The solutions to the equation r^2-4r-91=7 are r=2+sqrt(102),r=2-sqrt(102)
Find the zeros of r^2-4r-91=7
The zeros of r^2-4r-91=7 are r=2+sqrt(102),r=2-sqrt(102)