What are the solutions to the equation 0=-2x^2+40x-72 ?

The solutions to the equation 0=-2x^2+40x-72 are x=2,x=18

Find the zeros of 0=-2x^2+40x-72

The zeros of 0=-2x^2+40x-72 are x=2,x=18

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0 = - 2 x^{2} + 40 x - 72

x = 2, x = 18

Solution steps

0 = - 2 x^{2} + 40 x - 72

Switch sides

- 2 x^{2} + 40 x - 72 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 2 ) ( - 72 )}{2 ( - 2 )}

4 0^{2} - 4 (- 2) (- 72) = 32

x_{1, 2} = \frac{- 40 \pm 32}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 40 + 32}{2 ( - 2 )}, x_{2} = \frac{- 40 - 32}{2 ( - 2 )}

x = \frac{- 40 + 32}{2 ( - 2 )} : 2

x = \frac{- 40 - 32}{2 ( - 2 )} : 18

The solutions to the quadratic equation are:

x = 2, x = 18

0.9 x^{2} - 10 x - 28 = 0

x (2 x - 3) = x

81 x^{2} + 196 = 0

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

2 m^{2} + 7 m - 9 = 0

What are the solutions to the equation 0=-2x^2+40x-72 ?
The solutions to the equation 0=-2x^2+40x-72 are x=2,x=18
Find the zeros of 0=-2x^2+40x-72
The zeros of 0=-2x^2+40x-72 are x=2,x=18