What are the solutions to the equation (x-2)^2+x(x+1)=24 ?

The solutions to the equation (x-2)^2+x(x+1)=24 are x=4,x=-5/2

Find the zeros of (x-2)^2+x(x+1)=24

The zeros of (x-2)^2+x(x+1)=24 are x=4,x=-5/2

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(x - 2)^{2} + x (x + 1) = 24

x = 4, x = - \frac{5}{2}

Decimal Notation

x = 4, x = - 2.5

Solution steps

(x - 2)^{2} + x (x + 1) = 24

Expand

(x - 2)^{2} + x (x + 1) : 2 x^{2} - 3 x + 4

2 x^{2} - 3 x + 4 = 24

Move

24

to the left side

2 x^{2} - 3 x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 ( - 20 )}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 (- 20) = 13

x_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 13}{2 \cdot 2}, x_{2} = \frac{- ( - 3 ) - 13}{2 \cdot 2}

x = \frac{- ( - 3 ) + 13}{2 \cdot 2} : 4

x = \frac{- ( - 3 ) - 13}{2 \cdot 2} : - \frac{5}{2}

The solutions to the quadratic equation are:

x = 4, x = - \frac{5}{2}

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

3 x^{2} + 12 x - 15 = 0

3 y^{2} - 33 y = - 54

1 x^{2} + 2 x - 3 = 0

- 3 x^{2} + 4 x + 15 = 0

What are the solutions to the equation (x-2)^2+x(x+1)=24 ?
The solutions to the equation (x-2)^2+x(x+1)=24 are x=4,x=-5/2
Find the zeros of (x-2)^2+x(x+1)=24
The zeros of (x-2)^2+x(x+1)=24 are x=4,x=-5/2