What are the solutions to the equation x^2-3x-4=2x ?

The solutions to the equation x^2-3x-4=2x are x=(5+sqrt(41))/2 ,x=(5-sqrt(41))/2

Find the zeros of x^2-3x-4=2x

The zeros of x^2-3x-4=2x are x=(5+sqrt(41))/2 ,x=(5-sqrt(41))/2

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x^{2} - 3 x - 4 = 2 x

x = \frac{5 + 41}{2}, x = \frac{5 - 41}{2}

Solution steps

x^{2} - 3 x - 4 = 2 x

Move

2 x

to the left side

x^{2} - 5 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 4 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 4) = 41

x_{1, 2} = \frac{- ( - 5 ) \pm 41}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 41}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 41}{2 \cdot 1}

x = \frac{- ( - 5 ) + 41}{2 \cdot 1} : \frac{5 + 41}{2}

x = \frac{- ( - 5 ) - 41}{2 \cdot 1} : \frac{5 - 41}{2}

The solutions to the quadratic equation are:

x = \frac{5 + 41}{2}, x = \frac{5 - 41}{2}

x, x + y = x^{2} - y^{2}

(4 x + 7)^{2} = 44

35 - 6 x = 8 x^{2}

30 (p^{2} - 1) = 11 p

2 x^{2} = 10 x - 25

What are the solutions to the equation x^2-3x-4=2x ?
The solutions to the equation x^2-3x-4=2x are x=(5+sqrt(41))/2 ,x=(5-sqrt(41))/2
Find the zeros of x^2-3x-4=2x
The zeros of x^2-3x-4=2x are x=(5+sqrt(41))/2 ,x=(5-sqrt(41))/2