What are the solutions to the equation x^2-5x-3=2x-5 ?

The solutions to the equation x^2-5x-3=2x-5 are x=(7+sqrt(41))/2 ,x=(7-sqrt(41))/2

Find the zeros of x^2-5x-3=2x-5

The zeros of x^2-5x-3=2x-5 are x=(7+sqrt(41))/2 ,x=(7-sqrt(41))/2

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x^{2} - 5 x - 3 = 2 x - 5

x = \frac{7 + 41}{2}, x = \frac{7 - 41}{2}

Solution steps

x^{2} - 5 x - 3 = 2 x - 5

Move

5

to the left side

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

Move

2 x

to the left side

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

Solve with the quadratic formula

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

(- 7)^{2} - 4 \cdot 1 \cdot 2 = 41

x_{1, 2} = \frac{- ( - 7 ) \pm 41}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 41}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 41}{2 \cdot 1}

x = \frac{- ( - 7 ) + 41}{2 \cdot 1} : \frac{7 + 41}{2}

x = \frac{- ( - 7 ) - 41}{2 \cdot 1} : \frac{7 - 41}{2}

The solutions to the quadratic equation are:

x = \frac{7 + 41}{2}, x = \frac{7 - 41}{2}

- 19.6 (\frac{45}{196})^{2} + 9 (\frac{45}{196}) + 3

(x - \frac{1}{4})^{2} = \frac{9}{16}

(\frac{9}{4})^{\frac{5}{2}}

\frac{1}{3} + \frac{2}{5}

What are the solutions to the equation x^2-5x-3=2x-5 ?
The solutions to the equation x^2-5x-3=2x-5 are x=(7+sqrt(41))/2 ,x=(7-sqrt(41))/2
Find the zeros of x^2-5x-3=2x-5
The zeros of x^2-5x-3=2x-5 are x=(7+sqrt(41))/2 ,x=(7-sqrt(41))/2