What are the solutions to the equation a^2+4(3+a)a=5 ?

The solutions to the equation a^2+4(3+a)a=5 are a=(-6+sqrt(61))/5 ,a=-(6+sqrt(61))/5

Find the zeros of a^2+4(3+a)a=5

The zeros of a^2+4(3+a)a=5 are a=(-6+sqrt(61))/5 ,a=-(6+sqrt(61))/5

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a^{2} + 4 (3 + a) a = 5

a = \frac{- 6 + 61}{5}, a = - \frac{6 + 61}{5}

Decimal Notation

a = 0.36204 \dots, a = - 2.76204 \dots

Solution steps

a^{2} + 4 (3 + a) a = 5

Expand

a^{2} + 4 (3 + a) a : 5 a^{2} + 12 a

5 a^{2} + 12 a = 5

Move

5

to the left side

5 a^{2} + 12 a - 5 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

1 2^{2} - 4 \cdot 5 (- 5) = 261

a_{1, 2} = \frac{- 12 \pm 2 61}{2 \cdot 5}

Separate the solutions

a_{1} = \frac{- 12 + 2 61}{2 \cdot 5}, a_{2} = \frac{- 12 - 2 61}{2 \cdot 5}

a = \frac{- 12 + 2 61}{2 \cdot 5} : \frac{- 6 + 61}{5}

a = \frac{- 12 - 2 61}{2 \cdot 5} : - \frac{6 + 61}{5}

The solutions to the quadratic equation are:

a = \frac{- 6 + 61}{5}, a = - \frac{6 + 61}{5}

3 - x^{2} = 1 - x

2 x^{2} + 8 x - 11 = 0

\frac{1}{3} (u - \frac{3}{2}) (u + \frac{3}{2}) = u (\frac{u}{3} \frac{1}{4})

2 x^{2} - 10 x = 0

x - 3 - (- x (x + 1) + 4) - 5 = (- x + 6)

What are the solutions to the equation a^2+4(3+a)a=5 ?
The solutions to the equation a^2+4(3+a)a=5 are a=(-6+sqrt(61))/5 ,a=-(6+sqrt(61))/5
Find the zeros of a^2+4(3+a)a=5
The zeros of a^2+4(3+a)a=5 are a=(-6+sqrt(61))/5 ,a=-(6+sqrt(61))/5