What are the solutions to the equation x(x-5)=7-2x ?

The solutions to the equation x(x-5)=7-2x are x=(3+sqrt(37))/2 ,x=(3-sqrt(37))/2

Find the zeros of x(x-5)=7-2x

The zeros of x(x-5)=7-2x are x=(3+sqrt(37))/2 ,x=(3-sqrt(37))/2

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x (x - 5) = 7 - 2 x

x = \frac{3 + 37}{2}, x = \frac{3 - 37}{2}

Decimal Notation

x = 4.54138 \dots, x = - 1.54138 \dots

Solution steps

x (x - 5) = 7 - 2 x

Expand

x (x - 5) : x^{2} - 5 x

x^{2} - 5 x = 7 - 2 x

Move

2 x

to the left side

x^{2} - 3 x = 7

Move

7

to the left side

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 7) = 37

x_{1, 2} = \frac{- ( - 3 ) \pm 37}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 37}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 37}{2 \cdot 1}

x = \frac{- ( - 3 ) + 37}{2 \cdot 1} : \frac{3 + 37}{2}

x = \frac{- ( - 3 ) - 37}{2 \cdot 1} : \frac{3 - 37}{2}

The solutions to the quadratic equation are:

x = \frac{3 + 37}{2}, x = \frac{3 - 37}{2}

(x + 19)^{2} = - 36

5 m^{2} + 4 = m

2 w^{2} + 3 w - 54 = 0

x, x^{2} = ln (y) - ln (3)

k^{2} - 5 k = 6

What are the solutions to the equation x(x-5)=7-2x ?
The solutions to the equation x(x-5)=7-2x are x=(3+sqrt(37))/2 ,x=(3-sqrt(37))/2
Find the zeros of x(x-5)=7-2x
The zeros of x(x-5)=7-2x are x=(3+sqrt(37))/2 ,x=(3-sqrt(37))/2