What are the solutions to the equation 2x^2-2x-180=0 ?

The solutions to the equation 2x^2-2x-180=0 are x=10,x=-9

Find the zeros of 2x^2-2x-180=0

The zeros of 2x^2-2x-180=0 are x=10,x=-9

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2 x^{2} - 2 x - 180 = 0

x = 10, x = - 9

Solution steps

2 x^{2} - 2 x - 180 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 ( - 180 )}{2 \cdot 2}

(- 2)^{2} - 4 \cdot 2 (- 180) = 38

x_{1, 2} = \frac{- ( - 2 ) \pm 38}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 38}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 38}{2 \cdot 2}

x = \frac{- ( - 2 ) + 38}{2 \cdot 2} : 10

x = \frac{- ( - 2 ) - 38}{2 \cdot 2} : - 9

The solutions to the quadratic equation are:

x = 10, x = - 9

- 2 (3) + 4 (5) > 5

2 x (x^{2} - 4) > 0

x^{2} + 30 x + 56

21 x y - 9 x - 56 y^{2} + 24 y

4 x^{2} + 4 x - 35

What are the solutions to the equation 2x^2-2x-180=0 ?
The solutions to the equation 2x^2-2x-180=0 are x=10,x=-9
Find the zeros of 2x^2-2x-180=0
The zeros of 2x^2-2x-180=0 are x=10,x=-9