What are the solutions to the equation 7x^2+28x-35=0 ?

The solutions to the equation 7x^2+28x-35=0 are x=1,x=-5

Find the zeros of 7x^2+28x-35=0

The zeros of 7x^2+28x-35=0 are x=1,x=-5

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7 x^{2} + 28 x - 35 = 0

x = 1, x = - 5

Solution steps

7 x^{2} + 28 x - 35 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 7 ( - 35 )}{2 \cdot 7}

2 8^{2} - 4 \cdot 7 (- 35) = 42

x_{1, 2} = \frac{- 28 \pm 42}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- 28 + 42}{2 \cdot 7}, x_{2} = \frac{- 28 - 42}{2 \cdot 7}

x = \frac{- 28 + 42}{2 \cdot 7} : 1

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

The solutions to the quadratic equation are:

x = 1, x = - 5

- 5 m^{2} + 12 m + 8 = 3 - 5 m

x^{2} - 4 x = - 5 x

z (z - 3) = 5

6 x^{2} - 21 = - 5 x

5 n^{2} - 30 n + 20 = - n

What are the solutions to the equation 7x^2+28x-35=0 ?
The solutions to the equation 7x^2+28x-35=0 are x=1,x=-5
Find the zeros of 7x^2+28x-35=0
The zeros of 7x^2+28x-35=0 are x=1,x=-5