What are the solutions to the equation x^2+5x+9/4 =0 ?

The solutions to the equation x^2+5x+9/4 =0 are x=-1/2 ,x=-9/2

Find the zeros of x^2+5x+9/4 =0

The zeros of x^2+5x+9/4 =0 are x=-1/2 ,x=-9/2

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x^{2} + 5 x + \frac{9}{4} = 0

x = - \frac{1}{2}, x = - \frac{9}{2}

Decimal Notation

x = - 0.5, x = - 4.5

Solution steps

x^{2} + 5 x + \frac{9}{4} = 0

Multiply both sides by

4

4 x^{2} + 20 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 9}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 \cdot 9 = 16

x_{1, 2} = \frac{- 20 \pm 16}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 16}{2 \cdot 4}, x_{2} = \frac{- 20 - 16}{2 \cdot 4}

x = \frac{- 20 + 16}{2 \cdot 4} : - \frac{1}{2}

x = \frac{- 20 - 16}{2 \cdot 4} : - \frac{9}{2}

The solutions to the quadratic equation are:

x = - \frac{1}{2}, x = - \frac{9}{2}

\frac{1}{x + 1} + \frac{1}{x + 2} \geq 0

- 3 < 3 x + 7 \leq 4

2^{2} + x^{2} = 5^{2}

\frac{2}{7} y + 5 = - 9

\frac{5}{6} \cdot \frac{1}{4}

What are the solutions to the equation x^2+5x+9/4 =0 ?
The solutions to the equation x^2+5x+9/4 =0 are x=-1/2 ,x=-9/2
Find the zeros of x^2+5x+9/4 =0
The zeros of x^2+5x+9/4 =0 are x=-1/2 ,x=-9/2