What are the solutions to the equation 62= 1/2 (h-3)h ?

The solutions to the equation 62= 1/2 (h-3)h are h=(3+sqrt(505))/2 ,h=(3-sqrt(505))/2

Find the zeros of 62= 1/2 (h-3)h

The zeros of 62= 1/2 (h-3)h are h=(3+sqrt(505))/2 ,h=(3-sqrt(505))/2

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

h = \frac{3 + 505}{2}, h = \frac{3 - 505}{2}

Decimal Notation

h = 12.73610 \dots, h = - 9.73610 \dots

Solution steps

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

Multiply both sides by

2

124 = h (h - 3)

Expand

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

124 = h^{2} - 3 h

Switch sides

h^{2} - 3 h = 124

Move

124

to the left side

h^{2} - 3 h - 124 = 0

Solve with the quadratic formula

h_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 124 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 124) = 505

h_{1, 2} = \frac{- ( - 3 ) \pm 505}{2 \cdot 1}

Separate the solutions

h_{1} = \frac{- ( - 3 ) + 505}{2 \cdot 1}, h_{2} = \frac{- ( - 3 ) - 505}{2 \cdot 1}

h = \frac{- ( - 3 ) + 505}{2 \cdot 1} : \frac{3 + 505}{2}

h = \frac{- ( - 3 ) - 505}{2 \cdot 1} : \frac{3 - 505}{2}

The solutions to the quadratic equation are:

h = \frac{3 + 505}{2}, h = \frac{3 - 505}{2}

10 x^{2} - 27 x - 9 = 0

6 x^{2} = 5 x - 4

\frac{3}{4} x^{2} - \frac{1}{3} = 5

x^{2} - 11 x + 24 = 0

6 (y - 8)^{2} = 18

What are the solutions to the equation 62= 1/2 (h-3)h ?
The solutions to the equation 62= 1/2 (h-3)h are h=(3+sqrt(505))/2 ,h=(3-sqrt(505))/2
Find the zeros of 62= 1/2 (h-3)h
The zeros of 62= 1/2 (h-3)h are h=(3+sqrt(505))/2 ,h=(3-sqrt(505))/2