What are the solutions to the equation h^2+48=16h ?

The solutions to the equation h^2+48=16h are h=12,h=4

Find the zeros of h^2+48=16h

The zeros of h^2+48=16h are h=12,h=4

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h^{2} + 48 = 16 h

h = 12, h = 4

Solution steps

h^{2} + 48 = 16 h

Move

16 h

to the left side

h^{2} + 48 - 16 h = 0

Write in the standard form

a x^{2} + b x + c = 0

h^{2} - 16 h + 48 = 0

Solve with the quadratic formula

h_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot 48}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot 48 = 8

h_{1, 2} = \frac{- ( - 16 ) \pm 8}{2 \cdot 1}

Separate the solutions

h_{1} = \frac{- ( - 16 ) + 8}{2 \cdot 1}, h_{2} = \frac{- ( - 16 ) - 8}{2 \cdot 1}

h = \frac{- ( - 16 ) + 8}{2 \cdot 1} : 12

h = \frac{- ( - 16 ) - 8}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

h = 12, h = 4

2 a^{2} + 7 ab - 15 b^{2}

200 \cdot \frac{π}{12} \cdot \frac{1.16}{1.3} \cdot 5

2 i z + \frac{2 i}{1 - i} = 4 i

4 x + 3 = 11

(- 1)^{n} \cdot (- 1)^{n}

What are the solutions to the equation h^2+48=16h ?
The solutions to the equation h^2+48=16h are h=12,h=4
Find the zeros of h^2+48=16h
The zeros of h^2+48=16h are h=12,h=4