What are the solutions to the equation -x^2+4x+12=x+8 ?

The solutions to the equation -x^2+4x+12=x+8 are x=-1,x=4

Find the zeros of -x^2+4x+12=x+8

The zeros of -x^2+4x+12=x+8 are x=-1,x=4

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- x^{2} + 4 x + 12 = x + 8

x = - 1, x = 4

Solution steps

- x^{2} + 4 x + 12 = x + 8

Move

8

to the left side

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

Move

x

to the left side

- x^{2} + 3 x + 4 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1, x = 4

\frac{( 1 - ( \frac{1}{( 1 + 0.08 ) ^{3}} ) )}{0.08}

\frac{a ^{- 1} + b ^{- 1}}{a + b}

5 b + 3 = 28

12 (x + 3) = - 3 (x - 27)

9 m^{2} - 40 m^{2} + 16 = 0

What are the solutions to the equation -x^2+4x+12=x+8 ?
The solutions to the equation -x^2+4x+12=x+8 are x=-1,x=4
Find the zeros of -x^2+4x+12=x+8
The zeros of -x^2+4x+12=x+8 are x=-1,x=4