What are the solutions to the equation 29(x+1)+3=-x^2+27x+6 ?

The solutions to the equation 29(x+1)+3=-x^2+27x+6 are x=-1-5i,x=-1+5i

Find the zeros of 29(x+1)+3=-x^2+27x+6

The zeros of 29(x+1)+3=-x^2+27x+6 are x=-1-5i,x=-1+5i

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29 (x + 1) + 3 = - x^{2} + 27 x + 6

x = - 1 - 5 i, x = - 1 + 5 i

Solution steps

29 (x + 1) + 3 = - x^{2} + 27 x + 6

Expand

29 (x + 1) + 3 : 29 x + 32

29 x + 32 = - x^{2} + 27 x + 6

Switch sides

- x^{2} + 27 x + 6 = 29 x + 32

Move

32

to the left side

- x^{2} + 27 x - 26 = 29 x

Move

29 x

to the left side

- x^{2} - 2 x - 26 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 1 ) ( - 26 )}{2 ( - 1 )}

Simplify

(- 2)^{2} - 4 (- 1) (- 26) : 10 i

x_{1, 2} = \frac{- ( - 2 ) \pm 10 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 10 i}{2 ( - 1 )}, x_{2} = \frac{- ( - 2 ) - 10 i}{2 ( - 1 )}

x = \frac{- ( - 2 ) + 10 i}{2 ( - 1 )} : - 1 - 5 i

x = \frac{- ( - 2 ) - 10 i}{2 ( - 1 )} : - 1 + 5 i

The solutions to the quadratic equation are:

x = - 1 - 5 i, x = - 1 + 5 i

\frac{350 x ^{2}}{2} = 6.64 \cdot 1257 \cdot (540 - x)

2 (3 h - 1) = 4 - 2 h (h + 5)

(2 x - 3)^{2} = (x + 1) (2 x - 3)

4 x^{2} - 14 x + 2 = 0

0.3 q^{2} + q - 40 = 0

What are the solutions to the equation 29(x+1)+3=-x^2+27x+6 ?
The solutions to the equation 29(x+1)+3=-x^2+27x+6 are x=-1-5i,x=-1+5i
Find the zeros of 29(x+1)+3=-x^2+27x+6
The zeros of 29(x+1)+3=-x^2+27x+6 are x=-1-5i,x=-1+5i