What are the solutions to the equation -h(h-3)+1=4h-1 ?

The solutions to the equation -h(h-3)+1=4h-1 are h=-2,h=1

Find the zeros of -h(h-3)+1=4h-1

The zeros of -h(h-3)+1=4h-1 are h=-2,h=1

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- h (h - 3) + 1 = 4 h - 1

h = - 2, h = 1

Solution steps

- h (h - 3) + 1 = 4 h - 1

Expand

- h (h - 3) + 1 : - h^{2} + 3 h + 1

- h^{2} + 3 h + 1 = 4 h - 1

Move

1

to the left side

- h^{2} + 3 h + 2 = 4 h

Move

4 h

to the left side

- h^{2} - h + 2 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 (- 1) \cdot 2 = 3

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

Separate the solutions

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

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

h = \frac{- ( - 1 ) - 3}{2 ( - 1 )} : 1

The solutions to the quadratic equation are:

h = - 2, h = 1

\frac{5}{2} (x - 2) + 2 x - 4 = 9

3 (x + 4)^{2} = 10 x + 32

\frac{x}{5} = \frac{10}{13}

4 y^{2} + 9 y + 5 = 0

u^{2} + 10 u + 25 = 0

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