What are the solutions to the equation 2x^2+(x-2)(x+5)=(x+3)(x-4) ?

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

Find the zeros of 2x^2+(x-2)(x+5)=(x+3)(x-4)

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

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2 x^{2} + (x - 2) (x + 5) = (x + 3) (x - 4)

x = - 1

Solution steps

2 x^{2} + (x - 2) (x + 5) = (x + 3) (x - 4)

Expand

2 x^{2} + (x - 2) (x + 5) : 3 x^{2} + 3 x - 10

Expand

(x + 3) (x - 4) : x^{2} - x - 12

3 x^{2} + 3 x - 10 = x^{2} - x - 12

Move

12

to the left side

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

Move

x

to the left side

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

Move

x^{2}

to the left side

2 x^{2} + 4 x + 2 = 0

Solve with the quadratic formula

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

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

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

x = \frac{- 4}{2 \cdot 2}

\frac{- 4}{2 \cdot 2} = - 1

x = - 1

The solution to the quadratic equation is:

x = - 1

x^{2} + x + 56 = 0

5 x^{2} + 11 = 7

m^{2} + 4 m - 32 = 0

x^{2} + 32 x + 16 = 0

y^{2} - 20 y + 100 = 0

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