What are the solutions to the equation 4m^2+8m-4=1 ?

The solutions to the equation 4m^2+8m-4=1 are m= 1/2 ,m=-5/2

Find the zeros of 4m^2+8m-4=1

The zeros of 4m^2+8m-4=1 are m= 1/2 ,m=-5/2

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4 m^{2} + 8 m - 4 = 1

m = \frac{1}{2}, m = - \frac{5}{2}

Decimal Notation

m = 0.5, m = - 2.5

Solution steps

4 m^{2} + 8 m - 4 = 1

Move

1

to the left side

4 m^{2} + 8 m - 5 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

8^{2} - 4 \cdot 4 (- 5) = 12

m_{1, 2} = \frac{- 8 \pm 12}{2 \cdot 4}

Separate the solutions

m_{1} = \frac{- 8 + 12}{2 \cdot 4}, m_{2} = \frac{- 8 - 12}{2 \cdot 4}

m = \frac{- 8 + 12}{2 \cdot 4} : \frac{1}{2}

m = \frac{- 8 - 12}{2 \cdot 4} : - \frac{5}{2}

The solutions to the quadratic equation are:

m = \frac{1}{2}, m = - \frac{5}{2}

38 a^{4} b^{3} c^{2} + 12 a^{9} b^{8} - 34 a^{5} b c^{6}

9 (\frac{2}{3})^{1}

\frac{3}{4} (x + 8) = 9

\frac{\frac{x}{y} - \frac{y}{x}}{\frac{1}{5 x ^{2}} - \frac{1}{5 y ^{2}}}

25 c^{2} - 5 c - 16 = - 4

What are the solutions to the equation 4m^2+8m-4=1 ?
The solutions to the equation 4m^2+8m-4=1 are m= 1/2 ,m=-5/2
Find the zeros of 4m^2+8m-4=1
The zeros of 4m^2+8m-4=1 are m= 1/2 ,m=-5/2