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

The solutions to the equation x^2+7x=4 are x=(-7+sqrt(65))/2 ,x=(-7-sqrt(65))/2

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

The zeros of x^2+7x=4 are x=(-7+sqrt(65))/2 ,x=(-7-sqrt(65))/2

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x^{2} + 7 x = 4

x = \frac{- 7 + 65}{2}, x = \frac{- 7 - 65}{2}

Decimal Notation

x = 0.53112 \dots, x = - 7.53112 \dots

Solution steps

x^{2} + 7 x = 4

Move

4

to the left side

x^{2} + 7 x - 4 = 0

Solve with the quadratic formula

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

7^{2} - 4 \cdot 1 \cdot (- 4) = 65

x_{1, 2} = \frac{- 7 \pm 65}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 65}{2 \cdot 1}, x_{2} = \frac{- 7 - 65}{2 \cdot 1}

x = \frac{- 7 + 65}{2 \cdot 1} : \frac{- 7 + 65}{2}

x = \frac{- 7 - 65}{2 \cdot 1} : \frac{- 7 - 65}{2}

The solutions to the quadratic equation are:

x = \frac{- 7 + 65}{2}, x = \frac{- 7 - 65}{2}

5 x - 1 = 14

16 x^{2} = 80 x

25 n + 2 = 7

4 x^{2} - 5 x - 6

(\frac{( 12 x y ^{7} )}{4 x ^{4} y ^{3}})^{2}

What are the solutions to the equation x^2+7x=4 ?
The solutions to the equation x^2+7x=4 are x=(-7+sqrt(65))/2 ,x=(-7-sqrt(65))/2
Find the zeros of x^2+7x=4
The zeros of x^2+7x=4 are x=(-7+sqrt(65))/2 ,x=(-7-sqrt(65))/2