What are the solutions to the equation x^2+7x=3 ?

The solutions to the equation x^2+7x=3 are x=(-7+sqrt(61))/2 ,x=(-7-sqrt(61))/2

Find the zeros of x^2+7x=3

The zeros of x^2+7x=3 are x=(-7+sqrt(61))/2 ,x=(-7-sqrt(61))/2

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x^{2} + 7 x = 3

x = \frac{- 7 + 61}{2}, x = \frac{- 7 - 61}{2}

Solution steps

x^{2} + 7 x = 3

Move

3

to the left side

x^{2} + 7 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 3) = 61

x_{1, 2} = \frac{- 7 \pm 61}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 61}{2 \cdot 1}, x_{2} = \frac{- 7 - 61}{2 \cdot 1}

x = \frac{- 7 + 61}{2 \cdot 1} : \frac{- 7 + 61}{2}

x = \frac{- 7 - 61}{2 \cdot 1} : \frac{- 7 - 61}{2}

The solutions to the quadratic equation are:

x = \frac{- 7 + 61}{2}, x = \frac{- 7 - 61}{2}

4 (x + 3) - 7 > - 2 (x - 5) + 7

x \cdot 1 - x

- 4 x > 8

\frac{1 2 ^{- 1}}{( - 4 ) ^{- 1}}

x^{3} y - 4 x y

What are the solutions to the equation x^2+7x=3 ?
The solutions to the equation x^2+7x=3 are x=(-7+sqrt(61))/2 ,x=(-7-sqrt(61))/2
Find the zeros of x^2+7x=3
The zeros of x^2+7x=3 are x=(-7+sqrt(61))/2 ,x=(-7-sqrt(61))/2