What are the solutions to the equation 2x(x+5)=4 ?

The solutions to the equation 2x(x+5)=4 are x=(-5+sqrt(33))/2 ,x=-(5+sqrt(33))/2

Find the zeros of 2x(x+5)=4

The zeros of 2x(x+5)=4 are x=(-5+sqrt(33))/2 ,x=-(5+sqrt(33))/2

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2 x (x + 5) = 4

x = \frac{- 5 + 33}{2}, x = - \frac{5 + 33}{2}

Decimal Notation

x = 0.37228 \dots, x = - 5.37228 \dots

Solution steps

2 x (x + 5) = 4

Expand

2 x (x + 5) : 2 x^{2} + 10 x

2 x^{2} + 10 x = 4

Move

4

to the left side

2 x^{2} + 10 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

1 0^{2} - 4 \cdot 2 (- 4) = 233

x_{1, 2} = \frac{- 10 \pm 2 33}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 10 + 2 33}{2 \cdot 2}, x_{2} = \frac{- 10 - 2 33}{2 \cdot 2}

x = \frac{- 10 + 2 33}{2 \cdot 2} : \frac{- 5 + 33}{2}

x = \frac{- 10 - 2 33}{2 \cdot 2} : - \frac{5 + 33}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 33}{2}, x = - \frac{5 + 33}{2}

12 x^{2} y - 18 x^{2} - 8 y + 12

lo g_{10} (\frac{1}{2})

\frac{4}{3} \cdot \frac{3}{4}

∣ 3 x - 5 ∣ < - 5

3 x + \frac{1}{2} = \frac{1}{3} x - 8

What are the solutions to the equation 2x(x+5)=4 ?
The solutions to the equation 2x(x+5)=4 are x=(-5+sqrt(33))/2 ,x=-(5+sqrt(33))/2
Find the zeros of 2x(x+5)=4
The zeros of 2x(x+5)=4 are x=(-5+sqrt(33))/2 ,x=-(5+sqrt(33))/2