What are the solutions to the equation 4(x^2+14x+49)-1=0 ?

The solutions to the equation 4(x^2+14x+49)-1=0 are x=-13/2 ,x=-15/2

Find the zeros of 4(x^2+14x+49)-1=0

The zeros of 4(x^2+14x+49)-1=0 are x=-13/2 ,x=-15/2

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4 (x^{2} + 14 x + 49) - 1 = 0

x = - \frac{13}{2}, x = - \frac{15}{2}

Decimal Notation

x = - 6.5, x = - 7.5

Solution steps

4 (x^{2} + 14 x + 49) - 1 = 0

Expand

4 (x^{2} + 14 x + 49) - 1 : 4 x^{2} + 56 x + 195

4 x^{2} + 56 x + 195 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 56 \pm 5 6 ^{2} - 4 \cdot 4 \cdot 195}{2 \cdot 4}

5 6^{2} - 4 \cdot 4 \cdot 195 = 4

x_{1, 2} = \frac{- 56 \pm 4}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 56 + 4}{2 \cdot 4}, x_{2} = \frac{- 56 - 4}{2 \cdot 4}

x = \frac{- 56 + 4}{2 \cdot 4} : - \frac{13}{2}

x = \frac{- 56 - 4}{2 \cdot 4} : - \frac{15}{2}

The solutions to the quadratic equation are:

x = - \frac{13}{2}, x = - \frac{15}{2}

\frac{5}{19} \cdot 6

(x x)^{2}

2 x + 3 \geq 13

- 3 x + 1 = - 9

\frac{\frac{5}{6}}{2}

What are the solutions to the equation 4(x^2+14x+49)-1=0 ?
The solutions to the equation 4(x^2+14x+49)-1=0 are x=-13/2 ,x=-15/2
Find the zeros of 4(x^2+14x+49)-1=0
The zeros of 4(x^2+14x+49)-1=0 are x=-13/2 ,x=-15/2