What are the solutions to the equation 4x(x+39)+9=0 ?

The solutions to the equation 4x(x+39)+9=0 are x=(3(2sqrt(42)-13))/2 ,x=-(3(13+2sqrt(42)))/2

Find the zeros of 4x(x+39)+9=0

The zeros of 4x(x+39)+9=0 are x=(3(2sqrt(42)-13))/2 ,x=-(3(13+2sqrt(42)))/2

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4 x (x + 39) + 9 = 0

x = \frac{3 ( 2 42 - 13 )}{2}, x = - \frac{3 ( 13 + 2 42 )}{2}

Decimal Notation

x = - 0.05777 \dots, x = - 38.94222 \dots

Solution steps

4 x (x + 39) + 9 = 0

Expand

4 x (x + 39) + 9 : 4 x^{2} + 156 x + 9

4 x^{2} + 156 x + 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 156 \pm 15 6 ^{2} - 4 \cdot 4 \cdot 9}{2 \cdot 4}

15 6^{2} - 4 \cdot 4 \cdot 9 = 2442

x_{1, 2} = \frac{- 156 \pm 24 42}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 156 + 24 42}{2 \cdot 4}, x_{2} = \frac{- 156 - 24 42}{2 \cdot 4}

x = \frac{- 156 + 24 42}{2 \cdot 4} : \frac{3 ( 2 42 - 13 )}{2}

x = \frac{- 156 - 24 42}{2 \cdot 4} : - \frac{3 ( 13 + 2 42 )}{2}

The solutions to the quadratic equation are:

x = \frac{3 ( 2 42 - 13 )}{2}, x = - \frac{3 ( 13 + 2 42 )}{2}

x^{2} + 6 x + 6.75 = 0

55 = 3 t + 4 t^{2}

- 6 x^{2} + 4 x - 7 = 0

- 3 + y^{2} = y - y^{2}

4 k^{2} + 4 k + 1 = 0

What are the solutions to the equation 4x(x+39)+9=0 ?
The solutions to the equation 4x(x+39)+9=0 are x=(3(2sqrt(42)-13))/2 ,x=-(3(13+2sqrt(42)))/2
Find the zeros of 4x(x+39)+9=0
The zeros of 4x(x+39)+9=0 are x=(3(2sqrt(42)-13))/2 ,x=-(3(13+2sqrt(42)))/2