What are the solutions to the equation -2r^2-12r+10=-12 ?

The solutions to the equation -2r^2-12r+10=-12 are r=-3-2sqrt(5),r=2sqrt(5)-3

Find the zeros of -2r^2-12r+10=-12

The zeros of -2r^2-12r+10=-12 are r=-3-2sqrt(5),r=2sqrt(5)-3

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- 2 r^{2} - 12 r + 10 = - 12

r = - 3 - 25, r = 25 - 3

Solution steps

- 2 r^{2} - 12 r + 10 = - 12

Move

12

to the left side

- 2 r^{2} - 12 r + 22 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 2 ) \cdot 22}{2 ( - 2 )}

(- 12)^{2} - 4 (- 2) \cdot 22 = 85

r_{1, 2} = \frac{- ( - 12 ) \pm 8 5}{2 ( - 2 )}

Separate the solutions

r_{1} = \frac{- ( - 12 ) + 8 5}{2 ( - 2 )}, r_{2} = \frac{- ( - 12 ) - 8 5}{2 ( - 2 )}

r = \frac{- ( - 12 ) + 8 5}{2 ( - 2 )} : - 3 - 25

r = \frac{- ( - 12 ) - 8 5}{2 ( - 2 )} : 25 - 3

The solutions to the quadratic equation are:

r = - 3 - 25, r = 25 - 3

x^{2} + 4 x - 5 = 0

4 r^{2} + 44 r + 121 = 0

9 (x + 2)^{2} - 4 = 2

- 12 (x - 8) = 3 (x + 17)

125 - 300 n + 240 n^{2} - 64 n^{3}

What are the solutions to the equation -2r^2-12r+10=-12 ?
The solutions to the equation -2r^2-12r+10=-12 are r=-3-2sqrt(5),r=2sqrt(5)-3
Find the zeros of -2r^2-12r+10=-12
The zeros of -2r^2-12r+10=-12 are r=-3-2sqrt(5),r=2sqrt(5)-3