What are the solutions to the equation -(x+1)^2+3(x+1)+8=0 ?

The solutions to the equation -(x+1)^2+3(x+1)+8=0 are x=-(-1+sqrt(41))/2 ,x=(1+sqrt(41))/2

Find the zeros of -(x+1)^2+3(x+1)+8=0

The zeros of -(x+1)^2+3(x+1)+8=0 are x=-(-1+sqrt(41))/2 ,x=(1+sqrt(41))/2

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- (x + 1)^{2} + 3 (x + 1) + 8 = 0

x = - \frac{- 1 + 41}{2}, x = \frac{1 + 41}{2}

Decimal Notation

x = - 2.70156 \dots, x = 3.70156 \dots

Solution steps

- (x + 1)^{2} + 3 (x + 1) + 8 = 0

Expand

- (x + 1)^{2} + 3 (x + 1) + 8 : - x^{2} + x + 10

- x^{2} + x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 10}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 10 = 41

x_{1, 2} = \frac{- 1 \pm 41}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 1 + 41}{2 ( - 1 )}, x_{2} = \frac{- 1 - 41}{2 ( - 1 )}

x = \frac{- 1 + 41}{2 ( - 1 )} : - \frac{- 1 + 41}{2}

x = \frac{- 1 - 41}{2 ( - 1 )} : \frac{1 + 41}{2}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 41}{2}, x = \frac{1 + 41}{2}

x, y = 6 x - x^{2}

x^{2} + 9 x + 25 = 5

d^{2} = 5^{2} - 3^{2}

(m - 3) x^{2} + 4 m - 3 x = m x - 1

(3 w + 4) (2 w - 7) = 0

What are the solutions to the equation -(x+1)^2+3(x+1)+8=0 ?
The solutions to the equation -(x+1)^2+3(x+1)+8=0 are x=-(-1+sqrt(41))/2 ,x=(1+sqrt(41))/2
Find the zeros of -(x+1)^2+3(x+1)+8=0
The zeros of -(x+1)^2+3(x+1)+8=0 are x=-(-1+sqrt(41))/2 ,x=(1+sqrt(41))/2