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

The solutions to the equation x^2-5=(x+5)(2x-1) are x=0,x=-9

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

The zeros of x^2-5=(x+5)(2x-1) are x=0,x=-9

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

x = 0, x = - 9

Solution steps

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

Expand

(x + 5) (2 x - 1) : 2 x^{2} + 9 x - 5

x^{2} - 5 = 2 x^{2} + 9 x - 5

Switch sides

2 x^{2} + 9 x - 5 = x^{2} - 5

Move

5

to the left side

2 x^{2} + 9 x = x^{2}

Move

x^{2}

to the left side

x^{2} + 9 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

9^{2} - 4 \cdot 1 \cdot 0 = 9

x_{1, 2} = \frac{- 9 \pm 9}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 9 + 9}{2 \cdot 1}, x_{2} = \frac{- 9 - 9}{2 \cdot 1}

x = \frac{- 9 + 9}{2 \cdot 1} : 0

x = \frac{- 9 - 9}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

x = 0, x = - 9

9^{0}

6 x^{2} - 16 = 5 x^{2} - 6 x

ma - mb + na - nb + p a - p b

3 x^{3} - 9 x^{2} - 54 x

4 ∣ 5 - 2 h ∣ - 9 > 11

What are the solutions to the equation x^2-5=(x+5)(2x-1) ?
The solutions to the equation x^2-5=(x+5)(2x-1) are x=0,x=-9
Find the zeros of x^2-5=(x+5)(2x-1)
The zeros of x^2-5=(x+5)(2x-1) are x=0,x=-9