What are the solutions to the equation 2x(x+15)=15 ?

The solutions to the equation 2x(x+15)=15 are x=(-15+sqrt(255))/2 ,x=-(15+sqrt(255))/2

Find the zeros of 2x(x+15)=15

The zeros of 2x(x+15)=15 are x=(-15+sqrt(255))/2 ,x=-(15+sqrt(255))/2

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2 x (x + 15) = 15

x = \frac{- 15 + 255}{2}, x = - \frac{15 + 255}{2}

Decimal Notation

x = 0.48435 \dots, x = - 15.48435 \dots

Solution steps

2 x (x + 15) = 15

Expand

2 x (x + 15) : 2 x^{2} + 30 x

2 x^{2} + 30 x = 15

Move

15

to the left side

2 x^{2} + 30 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 30 \pm 3 0 ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

3 0^{2} - 4 \cdot 2 (- 15) = 2255

x_{1, 2} = \frac{- 30 \pm 2 255}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 30 + 2 255}{2 \cdot 2}, x_{2} = \frac{- 30 - 2 255}{2 \cdot 2}

x = \frac{- 30 + 2 255}{2 \cdot 2} : \frac{- 15 + 255}{2}

x = \frac{- 30 - 2 255}{2 \cdot 2} : - \frac{15 + 255}{2}

The solutions to the quadratic equation are:

x = \frac{- 15 + 255}{2}, x = - \frac{15 + 255}{2}

ln^{2} (e^{2})

41 \leq - 3 n - 1

2 x^{2} + 25 \leq x (x + 10)

(- m^{2} + 6) + (- 4 m^{2} + 7 m + 2)

\frac{a}{3} + \frac{a}{7} = \frac{20}{21}

What are the solutions to the equation 2x(x+15)=15 ?
The solutions to the equation 2x(x+15)=15 are x=(-15+sqrt(255))/2 ,x=-(15+sqrt(255))/2
Find the zeros of 2x(x+15)=15
The zeros of 2x(x+15)=15 are x=(-15+sqrt(255))/2 ,x=-(15+sqrt(255))/2