What are the solutions to the equation x^2+7x+49/4 =4 ?

The solutions to the equation x^2+7x+49/4 =4 are x=-3/2 ,x=-11/2

Find the zeros of x^2+7x+49/4 =4

The zeros of x^2+7x+49/4 =4 are x=-3/2 ,x=-11/2

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x^{2} + 7 x + \frac{49}{4} = 4

x = - \frac{3}{2}, x = - \frac{11}{2}

Decimal Notation

x = - 1.5, x = - 5.5

Solution steps

x^{2} + 7 x + \frac{49}{4} = 4

Multiply both sides by

4

4 x^{2} + 28 x + 49 = 16

Move

16

to the left side

4 x^{2} + 28 x + 33 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 \cdot 33}{2 \cdot 4}

2 8^{2} - 4 \cdot 4 \cdot 33 = 16

x_{1, 2} = \frac{- 28 \pm 16}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 28 + 16}{2 \cdot 4}, x_{2} = \frac{- 28 - 16}{2 \cdot 4}

x = \frac{- 28 + 16}{2 \cdot 4} : - \frac{3}{2}

x = \frac{- 28 - 16}{2 \cdot 4} : - \frac{11}{2}

The solutions to the quadratic equation are:

x = - \frac{3}{2}, x = - \frac{11}{2}

(x + 5)^{2} = 10 x + 50

3 x (5 - x) = x (7 - x)

x (2 x + \frac{1}{3}) - 5 x^{2} = 3 x - 2

4 (x^{2} - 1) = 21

15 (2 a - 2) = 5 (a^{2} - 1)

What are the solutions to the equation x^2+7x+49/4 =4 ?
The solutions to the equation x^2+7x+49/4 =4 are x=-3/2 ,x=-11/2
Find the zeros of x^2+7x+49/4 =4
The zeros of x^2+7x+49/4 =4 are x=-3/2 ,x=-11/2