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 2 8 ^{2} - 4 \cdot 4 \cdot 49}{2 \cdot 4}

2 8^{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}

5 + x \geq 12

41 = 12 d - 7

5 z^{2} + 18 z + 9 = 0

r^{2} + r - 2 = 0

- 18 < - 2 p \leq 12

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