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

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

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

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

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

x = \frac{7}{2}

Decimal Notation

x = 3.5

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

x = \frac{7}{2}

The solution to the quadratic equation is:

x = \frac{7}{2}

w^{2} - 3 w = 10

\frac{2}{3} = \frac{x}{6}

6 x^{2} + 7 x - 5

- 4 x + 6 = - 30

(\frac{3}{7} m^{3} + \frac{1}{4} n^{3}) (\frac{3}{7} m^{3} - \frac{1}{4} n^{3})

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