What are the solutions to the equation 7x^2+53x+28=0 ?

The solutions to the equation 7x^2+53x+28=0 are x=-4/7 ,x=-7

Find the zeros of 7x^2+53x+28=0

The zeros of 7x^2+53x+28=0 are x=-4/7 ,x=-7

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7 x^{2} + 53 x + 28 = 0

x = - \frac{4}{7}, x = - 7

Decimal

x = - 0.57142 \dots, x = - 7

Solution steps

7 x^{2} + 53 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 53 \pm 5 3 ^{2} - 4 \cdot 7 \cdot 28}{2 \cdot 7}

5 3^{2} - 4 \cdot 7 \cdot 28 = 45

x_{1, 2} = \frac{- 53 \pm 45}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- 53 + 45}{2 \cdot 7}, x_{2} = \frac{- 53 - 45}{2 \cdot 7}

x = \frac{- 53 + 45}{2 \cdot 7} : - \frac{4}{7}

x = \frac{- 53 - 45}{2 \cdot 7} : - 7

The solutions to the quadratic equation are:

x = - \frac{4}{7}, x = - 7

y^{2} - 12 = 4

- 3 x^{2} + x + 3 = 0

2 x^{2} + 5 = \frac{418}{16}

10 (p^{2} - 1) = 21 p

- 16 t^{2} + 70 t + 1 = 41

What are the solutions to the equation 7x^2+53x+28=0 ?
The solutions to the equation 7x^2+53x+28=0 are x=-4/7 ,x=-7
Find the zeros of 7x^2+53x+28=0
The zeros of 7x^2+53x+28=0 are x=-4/7 ,x=-7