What are the solutions to the equation 2x^2+7x-18=x^2 ?

The solutions to the equation 2x^2+7x-18=x^2 are x=2,x=-9

Find the zeros of 2x^2+7x-18=x^2

The zeros of 2x^2+7x-18=x^2 are x=2,x=-9

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2 x^{2} + 7 x - 18 = x^{2}

x = 2, x = - 9

Solution steps

2 x^{2} + 7 x - 18 = x^{2}

Move

x^{2}

to the left side

x^{2} + 7 x - 18 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

7^{2} - 4 \cdot 1 \cdot (- 18) = 11

x_{1, 2} = \frac{- 7 \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 7 + 11}{2 \cdot 1}, x_{2} = \frac{- 7 - 11}{2 \cdot 1}

x = \frac{- 7 + 11}{2 \cdot 1} : 2

x = \frac{- 7 - 11}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

x = 2, x = - 9

2 0^{2} + 2 \cdot 32.2 = v^{2} + 50 (5 - 2)^{2}

x, 3.1396 (1 - x^{2}) = (π) (x)^{2}

- t^{2} + 27 t + 3 = - 2 t^{2} + 25 t + 27

- 6 x^{2} = 4

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

What are the solutions to the equation 2x^2+7x-18=x^2 ?
The solutions to the equation 2x^2+7x-18=x^2 are x=2,x=-9
Find the zeros of 2x^2+7x-18=x^2
The zeros of 2x^2+7x-18=x^2 are x=2,x=-9