What are the solutions to the equation -x^2+13=3x ?

The solutions to the equation -x^2+13=3x are x=-(3+sqrt(61))/2 ,x=(sqrt(61)-3)/2

Find the zeros of -x^2+13=3x

The zeros of -x^2+13=3x are x=-(3+sqrt(61))/2 ,x=(sqrt(61)-3)/2

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- x^{2} + 13 = 3 x

x = - \frac{3 + 61}{2}, x = \frac{61 - 3}{2}

Decimal Notation

x = - 5.40512 \dots, x = 2.40512 \dots

Solution steps

- x^{2} + 13 = 3 x

Move

3 x

to the left side

- x^{2} + 13 - 3 x = 0

Write in the standard form

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

- x^{2} - 3 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) \cdot 13}{2 ( - 1 )}

(- 3)^{2} - 4 (- 1) \cdot 13 = 61

x_{1, 2} = \frac{- ( - 3 ) \pm 61}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 61}{2 ( - 1 )}, x_{2} = \frac{- ( - 3 ) - 61}{2 ( - 1 )}

x = \frac{- ( - 3 ) + 61}{2 ( - 1 )} : - \frac{3 + 61}{2}

x = \frac{- ( - 3 ) - 61}{2 ( - 1 )} : \frac{61 - 3}{2}

The solutions to the quadratic equation are:

x = - \frac{3 + 61}{2}, x = \frac{61 - 3}{2}

9 x^{2} - 6 x - 11 = 0

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

(x^{2} - 8) 2 + x^{2} - 8 = 20

5 x^{2} + 18 x + 2 = 0

z^{2} = - 4

What are the solutions to the equation -x^2+13=3x ?
The solutions to the equation -x^2+13=3x are x=-(3+sqrt(61))/2 ,x=(sqrt(61)-3)/2
Find the zeros of -x^2+13=3x
The zeros of -x^2+13=3x are x=-(3+sqrt(61))/2 ,x=(sqrt(61)-3)/2