What are the solutions to the equation -12x+x^2=28 ?

The solutions to the equation -12x+x^2=28 are x=14,x=-2

Find the zeros of -12x+x^2=28

The zeros of -12x+x^2=28 are x=14,x=-2

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- 12 x + x^{2} = 28

x = 14, x = - 2

Solution steps

- 12 x + x^{2} = 28

Move

28

to the left side

- 12 x + x^{2} - 28 = 0

Write in the standard form

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

x^{2} - 12 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot (- 28) = 16

x_{1, 2} = \frac{- ( - 12 ) \pm 16}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 16}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 16}{2 \cdot 1}

x = \frac{- ( - 12 ) + 16}{2 \cdot 1} : 14

x = \frac{- ( - 12 ) - 16}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 14, x = - 2

(x - 2)^{2} = 50

2 x^{2} - 4 x + c = 0

4 x^{2} + 8 x = - 1

h^{2} + 10 h + 9 = 0

4 y^{2} = 4

What are the solutions to the equation -12x+x^2=28 ?
The solutions to the equation -12x+x^2=28 are x=14,x=-2
Find the zeros of -12x+x^2=28
The zeros of -12x+x^2=28 are x=14,x=-2