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}

Decimal Notation

x = 7.40512 \dots, x = - 0.40512 \dots

Solution steps

x^{2} = 7 x + 3

Move

3

to the left side

x^{2} - 3 = 7 x

Move

7 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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}

5 y - [2 (y + 1) + 2] = 7 y

52 - 6 (7 w - 3) \geq 8 (4 - 3 w)

6 4^{\frac{7}{12}}

- 3 (x - 4) + x = 2 x - 12

x^{2} - 40 x + 300 = 0

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