What are the solutions to the equation 2(x^2-3x)-x(x-6)=25 ?

The solutions to the equation 2(x^2-3x)-x(x-6)=25 are x=5,x=-5

Find the zeros of 2(x^2-3x)-x(x-6)=25

The zeros of 2(x^2-3x)-x(x-6)=25 are x=5,x=-5

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2 (x^{2} - 3 x) - x (x - 6) = 25

x = 5, x = - 5

Solution steps

2 (x^{2} - 3 x) - x (x - 6) = 25

Expand

2 (x^{2} - 3 x) - x (x - 6) : x^{2}

x^{2} = 25

Move

25

to the left side

x^{2} - 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - 25 )}{2 \cdot 1}

0^{2} - 4 \cdot 1 \cdot (- 25) = 10

x_{1, 2} = \frac{- 0 \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 0 + 10}{2 \cdot 1}, x_{2} = \frac{- 0 - 10}{2 \cdot 1}

x = \frac{- 0 + 10}{2 \cdot 1} : 5

x = \frac{- 0 - 10}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 5, x = - 5

d^{2} + 3 d - 7 = 0

x, f (x + 1) = - 3 x^{2} - 1

2.6 \cdot 9.8 (3.5 + x) = 0.5 \cdot 512 (x^{2})

(z + 1)^{2} = 2 z^{2}

- 6.5 q^{2} + 390 q - 2000 = 0

What are the solutions to the equation 2(x^2-3x)-x(x-6)=25 ?
The solutions to the equation 2(x^2-3x)-x(x-6)=25 are x=5,x=-5
Find the zeros of 2(x^2-3x)-x(x-6)=25
The zeros of 2(x^2-3x)-x(x-6)=25 are x=5,x=-5