What are the solutions to the equation -w^2+5w-4=0 ?

The solutions to the equation -w^2+5w-4=0 are w=1,w=4

Find the zeros of -w^2+5w-4=0

The zeros of -w^2+5w-4=0 are w=1,w=4

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- w^{2} + 5 w - 4 = 0

w = 1, w = 4

Solution steps

- w^{2} + 5 w - 4 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 1 ) ( - 4 )}{2 ( - 1 )}

5^{2} - 4 (- 1) (- 4) = 3

w_{1, 2} = \frac{- 5 \pm 3}{2 ( - 1 )}

Separate the solutions

w_{1} = \frac{- 5 + 3}{2 ( - 1 )}, w_{2} = \frac{- 5 - 3}{2 ( - 1 )}

w = \frac{- 5 + 3}{2 ( - 1 )} : 1

w = \frac{- 5 - 3}{2 ( - 1 )} : 4

The solutions to the quadratic equation are:

w = 1, w = 4

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

4 x^{2} + 24 x + 37 = 0

x^{2} + x + 16 = 0

6 x^{2} = 2 - x

y, 2 x^{2} y + \frac{x ^{3} - y ^{2}}{x} + 3 = 0

What are the solutions to the equation -w^2+5w-4=0 ?
The solutions to the equation -w^2+5w-4=0 are w=1,w=4
Find the zeros of -w^2+5w-4=0
The zeros of -w^2+5w-4=0 are w=1,w=4