What are the solutions to the equation (2x-1)^2-x(x-3)=5 ?

The solutions to the equation (2x-1)^2-x(x-3)=5 are x= 4/3 ,x=-1

Find the zeros of (2x-1)^2-x(x-3)=5

The zeros of (2x-1)^2-x(x-3)=5 are x= 4/3 ,x=-1

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(2 x - 1)^{2} - x (x - 3) = 5

x = \frac{4}{3}, x = - 1

Solution steps

(2 x - 1)^{2} - x (x - 3) = 5

Expand

(2 x - 1)^{2} - x (x - 3) : 3 x^{2} - x + 1

3 x^{2} - x + 1 = 5

Move

5

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 3}, x_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 3}

x = \frac{- ( - 1 ) + 7}{2 \cdot 3} : \frac{4}{3}

x = \frac{- ( - 1 ) - 7}{2 \cdot 3} : - 1

The solutions to the quadratic equation are:

x = \frac{4}{3}, x = - 1

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

4 (1 + 4 x) = x (x - 7)

15 x^{2} + 6 x - 14 = 0

x, x^{2} + y^{2} - 3 x y + 4 = 0

3 x (x - 2) = 10

What are the solutions to the equation (2x-1)^2-x(x-3)=5 ?
The solutions to the equation (2x-1)^2-x(x-3)=5 are x= 4/3 ,x=-1
Find the zeros of (2x-1)^2-x(x-3)=5
The zeros of (2x-1)^2-x(x-3)=5 are x= 4/3 ,x=-1