What are the solutions to the equation -3x(x+5)=-9 ?

The solutions to the equation -3x(x+5)=-9 are x=-(5+sqrt(37))/2 ,x=(sqrt(37)-5)/2

Find the zeros of -3x(x+5)=-9

The zeros of -3x(x+5)=-9 are x=-(5+sqrt(37))/2 ,x=(sqrt(37)-5)/2

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- 3 x (x + 5) = - 9

x = - \frac{5 + 37}{2}, x = \frac{37 - 5}{2}

Decimal Notation

x = - 5.54138 \dots, x = 0.54138 \dots

Solution steps

- 3 x (x + 5) = - 9

Expand

- 3 x (x + 5) : - 3 x^{2} - 15 x

- 3 x^{2} - 15 x = - 9

Move

9

to the left side

- 3 x^{2} - 15 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 ( - 3 ) \cdot 9}{2 ( - 3 )}

(- 15)^{2} - 4 (- 3) \cdot 9 = 337

x_{1, 2} = \frac{- ( - 15 ) \pm 3 37}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 15 ) + 3 37}{2 ( - 3 )}, x_{2} = \frac{- ( - 15 ) - 3 37}{2 ( - 3 )}

x = \frac{- ( - 15 ) + 3 37}{2 ( - 3 )} : - \frac{5 + 37}{2}

x = \frac{- ( - 15 ) - 3 37}{2 ( - 3 )} : \frac{37 - 5}{2}

The solutions to the quadratic equation are:

x = - \frac{5 + 37}{2}, x = \frac{37 - 5}{2}

8 d^{2} + 8 d + 2 = 0

s^{2} + 5 s + 4

12 x^{2} - 4 x = 5

- t^{2} + 3 t + 27 = 0

x^{2} - 14 x + 49 = 0

What are the solutions to the equation -3x(x+5)=-9 ?
The solutions to the equation -3x(x+5)=-9 are x=-(5+sqrt(37))/2 ,x=(sqrt(37)-5)/2
Find the zeros of -3x(x+5)=-9
The zeros of -3x(x+5)=-9 are x=-(5+sqrt(37))/2 ,x=(sqrt(37)-5)/2