What are the solutions to the equation 5x^2+28x=49 ?

The solutions to the equation 5x^2+28x=49 are x= 7/5 ,x=-7

Find the zeros of 5x^2+28x=49

The zeros of 5x^2+28x=49 are x= 7/5 ,x=-7

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5 x^{2} + 28 x = 49

x = \frac{7}{5}, x = - 7

Decimal Notation

x = 1.4, x = - 7

Solution steps

5 x^{2} + 28 x = 49

Move

49

to the left side

5 x^{2} + 28 x - 49 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 5 ( - 49 )}{2 \cdot 5}

2 8^{2} - 4 \cdot 5 (- 49) = 42

x_{1, 2} = \frac{- 28 \pm 42}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 28 + 42}{2 \cdot 5}, x_{2} = \frac{- 28 - 42}{2 \cdot 5}

x = \frac{- 28 + 42}{2 \cdot 5} : \frac{7}{5}

x = \frac{- 28 - 42}{2 \cdot 5} : - 7

The solutions to the quadratic equation are:

x = \frac{7}{5}, x = - 7

8 1^{\frac{1}{4}}

3 y^{2} - 7 = 20

- \frac{14}{- 4}

x^{2} - 2 x \leq 0

4 p^{2} - 12 p + 9 = 0

What are the solutions to the equation 5x^2+28x=49 ?
The solutions to the equation 5x^2+28x=49 are x= 7/5 ,x=-7
Find the zeros of 5x^2+28x=49
The zeros of 5x^2+28x=49 are x= 7/5 ,x=-7