What are the solutions to the equation (x-5)^2+x(x+3)=25 ?

The solutions to the equation (x-5)^2+x(x+3)=25 are x= 7/2 ,x=0

Find the zeros of (x-5)^2+x(x+3)=25

The zeros of (x-5)^2+x(x+3)=25 are x= 7/2 ,x=0

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(x - 5)^{2} + x (x + 3) = 25

x = \frac{7}{2}, x = 0

Solution steps

(x - 5)^{2} + x (x + 3) = 25

Expand

(x - 5)^{2} + x (x + 3) : 2 x^{2} - 7 x + 25

2 x^{2} - 7 x + 25 = 25

Move

25

to the left side

2 x^{2} - 7 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 0 = 7

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

Separate the solutions

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

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

x = \frac{- ( - 7 ) - 7}{2 \cdot 2} : 0

The solutions to the quadratic equation are:

x = \frac{7}{2}, x = 0

5 x^{2} - 12 x - 8 = 0

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

2 (x - 1)^{2} - 8 = 0

29 = x^{2} + 4

x, f = u_{1} (x^{2})

What are the solutions to the equation (x-5)^2+x(x+3)=25 ?
The solutions to the equation (x-5)^2+x(x+3)=25 are x= 7/2 ,x=0
Find the zeros of (x-5)^2+x(x+3)=25
The zeros of (x-5)^2+x(x+3)=25 are x= 7/2 ,x=0