What are the solutions to the equation x^2+11x+121/4 = 125/4 ?

The solutions to the equation x^2+11x+121/4 = 125/4 are x=(-11+5sqrt(5))/2 ,x=-(11+5sqrt(5))/2

Find the zeros of x^2+11x+121/4 = 125/4

The zeros of x^2+11x+121/4 = 125/4 are x=(-11+5sqrt(5))/2 ,x=-(11+5sqrt(5))/2

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x^{2} + 11 x + \frac{121}{4} = \frac{125}{4}

x = \frac{- 11 + 5 5}{2}, x = - \frac{11 + 5 5}{2}

Solution steps

x^{2} + 11 x + \frac{121}{4} = \frac{125}{4}

Multiply both sides by

4

4 x^{2} + 44 x + 121 = 125

Move

125

to the left side

4 x^{2} + 44 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 44 \pm 4 4 ^{2} - 4 \cdot 4 ( - 4 )}{2 \cdot 4}

4 4^{2} - 4 \cdot 4 (- 4) = 205

x_{1, 2} = \frac{- 44 \pm 20 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 44 + 20 5}{2 \cdot 4}, x_{2} = \frac{- 44 - 20 5}{2 \cdot 4}

x = \frac{- 44 + 20 5}{2 \cdot 4} : \frac{- 11 + 5 5}{2}

x = \frac{- 44 - 20 5}{2 \cdot 4} : - \frac{11 + 5 5}{2}

The solutions to the quadratic equation are:

x = \frac{- 11 + 5 5}{2}, x = - \frac{11 + 5 5}{2}

ab + a c

3 d > 9

6 x - 18 = 0

4 n - 3 = 5

\frac{1}{2} y + 2 > \frac{1}{3} y - 4

What are the solutions to the equation x^2+11x+121/4 = 125/4 ?
The solutions to the equation x^2+11x+121/4 = 125/4 are x=(-11+5sqrt(5))/2 ,x=-(11+5sqrt(5))/2
Find the zeros of x^2+11x+121/4 = 125/4
The zeros of x^2+11x+121/4 = 125/4 are x=(-11+5sqrt(5))/2 ,x=-(11+5sqrt(5))/2