What are the solutions to the equation (11m+2-4m^2)^2-4(m^2+2)(4m^2+26-22m)=0 ?

The solutions to the equation (11m+2-4m^2)^2-4(m^2+2)(4m^2+26-22m)=0 are m= 34/31 ,m=6

Find the zeros of (11m+2-4m^2)^2-4(m^2+2)(4m^2+26-22m)=0

The zeros of (11m+2-4m^2)^2-4(m^2+2)(4m^2+26-22m)=0 are m= 34/31 ,m=6

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(11 m + 2 - 4 m^{2})^{2} - 4 (m^{2} + 2) (4 m^{2} + 26 - 22 m) = 0

m = \frac{34}{31}, m = 6

Decimal Notation

m = 1.09677 \dots, m = 6

Solution steps

(11 m + 2 - 4 m^{2})^{2} - 4 (m^{2} + 2) (4 m^{2} + 26 - 22 m) = 0

Expand

(11 m + 2 - 4 m^{2})^{2} - 4 (m^{2} + 2) (4 m^{2} + 26 - 22 m) : - 31 m^{2} + 220 m - 204

- 31 m^{2} + 220 m - 204 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 220 \pm 22 0 ^{2} - 4 ( - 31 ) ( - 204 )}{2 ( - 31 )}

22 0^{2} - 4 (- 31) (- 204) = 152

m_{1, 2} = \frac{- 220 \pm 152}{2 ( - 31 )}

Separate the solutions

m_{1} = \frac{- 220 + 152}{2 ( - 31 )}, m_{2} = \frac{- 220 - 152}{2 ( - 31 )}

m = \frac{- 220 + 152}{2 ( - 31 )} : \frac{34}{31}

m = \frac{- 220 - 152}{2 ( - 31 )} : 6

The solutions to the quadratic equation are:

m = \frac{34}{31}, m = 6

x, x^{2} = \frac{4 R}{3} (2 R - \frac{4 R}{3})

u^{2} - 4 u = 12

a, (a - b)^{2} = 12

x^{2} = 0.9

x, x^{2} + y^{2} + z^{2} - 4 y - 6 z + 13 = 0

What are the solutions to the equation (11m+2-4m^2)^2-4(m^2+2)(4m^2+26-22m)=0 ?
The solutions to the equation (11m+2-4m^2)^2-4(m^2+2)(4m^2+26-22m)=0 are m= 34/31 ,m=6
Find the zeros of (11m+2-4m^2)^2-4(m^2+2)(4m^2+26-22m)=0
The zeros of (11m+2-4m^2)^2-4(m^2+2)(4m^2+26-22m)=0 are m= 34/31 ,m=6