What are the solutions to the equation 4m^2-28m+49=0 ?

The solutions to the equation 4m^2-28m+49=0 are m= 7/2

Find the zeros of 4m^2-28m+49=0

The zeros of 4m^2-28m+49=0 are m= 7/2

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4 m^{2} - 28 m + 49 = 0

m = \frac{7}{2}

Decimal Notation

m = 3.5

Solution steps

4 m^{2} - 28 m + 49 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- ( - 28 ) \pm ( - 28 ) ^{2} - 4 \cdot 4 \cdot 49}{2 \cdot 4}

(- 28)^{2} - 4 \cdot 4 \cdot 49 = 0

m_{1, 2} = \frac{- ( - 28 ) \pm 0}{2 \cdot 4}

m = \frac{- ( - 28 )}{2 \cdot 4}

\frac{- ( - 28 )}{2 \cdot 4} = \frac{7}{2}

m = \frac{7}{2}

The solution to the quadratic equation is:

m = \frac{7}{2}

2 x^{2} - 4 x - 4

20 x^{2} - 7 x - 6

∣ x - 4 ∣ + 6 = 9

\frac{1}{2} \cdot 14

x^{2} - 2 x + 8 \leq 0

What are the solutions to the equation 4m^2-28m+49=0 ?
The solutions to the equation 4m^2-28m+49=0 are m= 7/2
Find the zeros of 4m^2-28m+49=0
The zeros of 4m^2-28m+49=0 are m= 7/2