What are the solutions to the equation -(x+2)^2-3x+5=0 ?

The solutions to the equation -(x+2)^2-3x+5=0 are x=-(7+sqrt(53))/2 ,x=(sqrt(53)-7)/2

Find the zeros of -(x+2)^2-3x+5=0

The zeros of -(x+2)^2-3x+5=0 are x=-(7+sqrt(53))/2 ,x=(sqrt(53)-7)/2

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- (x + 2)^{2} - 3 x + 5 = 0

x = - \frac{7 + 53}{2}, x = \frac{53 - 7}{2}

Decimal

x = - 7.14005 \dots, x = 0.14005 \dots

Solution steps

- (x + 2)^{2} - 3 x + 5 = 0

Expand

- (x + 2)^{2} - 3 x + 5 : - x^{2} - 7 x + 1

- x^{2} - 7 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

(- 7)^{2} - 4 (- 1) \cdot 1 = 53

x_{1, 2} = \frac{- ( - 7 ) \pm 53}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 53}{2 ( - 1 )}, x_{2} = \frac{- ( - 7 ) - 53}{2 ( - 1 )}

x = \frac{- ( - 7 ) + 53}{2 ( - 1 )} : - \frac{7 + 53}{2}

x = \frac{- ( - 7 ) - 53}{2 ( - 1 )} : \frac{53 - 7}{2}

The solutions to the quadratic equation are:

x = - \frac{7 + 53}{2}, x = \frac{53 - 7}{2}

x^{2} + 7 = 5 x + 1

0 = 5 x^{2} + 25

x, x^{2} + y^{2} + z^{2} - yz = 1

2 x^{2} - x = 3

5 (2 x + 8)^{2} = 80

What are the solutions to the equation -(x+2)^2-3x+5=0 ?
The solutions to the equation -(x+2)^2-3x+5=0 are x=-(7+sqrt(53))/2 ,x=(sqrt(53)-7)/2
Find the zeros of -(x+2)^2-3x+5=0
The zeros of -(x+2)^2-3x+5=0 are x=-(7+sqrt(53))/2 ,x=(sqrt(53)-7)/2