What are the solutions to the equation 4m^2-m-3=-m^2 ?

The solutions to the equation 4m^2-m-3=-m^2 are m=(1+sqrt(61))/(10),m=(1-sqrt(61))/(10)

Find the zeros of 4m^2-m-3=-m^2

The zeros of 4m^2-m-3=-m^2 are m=(1+sqrt(61))/(10),m=(1-sqrt(61))/(10)

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4 m^{2} - m - 3 = - m^{2}

m = \frac{1 + 61}{10}, m = \frac{1 - 61}{10}

Decimal Notation

m = 0.88102 \dots, m = - 0.68102 \dots

Solution steps

4 m^{2} - m - 3 = - m^{2}

Move

m^{2}

to the left side

5 m^{2} - m - 3 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

(- 1)^{2} - 4 \cdot 5 (- 3) = 61

m_{1, 2} = \frac{- ( - 1 ) \pm 61}{2 \cdot 5}

Separate the solutions

m_{1} = \frac{- ( - 1 ) + 61}{2 \cdot 5}, m_{2} = \frac{- ( - 1 ) - 61}{2 \cdot 5}

m = \frac{- ( - 1 ) + 61}{2 \cdot 5} : \frac{1 + 61}{10}

m = \frac{- ( - 1 ) - 61}{2 \cdot 5} : \frac{1 - 61}{10}

The solutions to the quadratic equation are:

m = \frac{1 + 61}{10}, m = \frac{1 - 61}{10}

20 x^{2} - x = 30

x^{3} - 3 x + 2 = (x + a)^{2} (x + b)

(2 x + 3) (3 x - 1) = 0

4 x^{2} - 40 x + 109 = 0

x, x y^{2} + 4 x y + 3 x = x^{2} - 1

What are the solutions to the equation 4m^2-m-3=-m^2 ?
The solutions to the equation 4m^2-m-3=-m^2 are m=(1+sqrt(61))/(10),m=(1-sqrt(61))/(10)
Find the zeros of 4m^2-m-3=-m^2
The zeros of 4m^2-m-3=-m^2 are m=(1+sqrt(61))/(10),m=(1-sqrt(61))/(10)