What are the solutions to the equation -2a^2-2a-3=-4a^2 ?

The solutions to the equation -2a^2-2a-3=-4a^2 are a=(1+sqrt(7))/2 ,a=(1-sqrt(7))/2

Find the zeros of -2a^2-2a-3=-4a^2

The zeros of -2a^2-2a-3=-4a^2 are a=(1+sqrt(7))/2 ,a=(1-sqrt(7))/2

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- 2 a^{2} - 2 a - 3 = - 4 a^{2}

a = \frac{1 + 7}{2}, a = \frac{1 - 7}{2}

Decimal Notation

a = 1.82287 \dots, a = - 0.82287 \dots

Solution steps

- 2 a^{2} - 2 a - 3 = - 4 a^{2}

Move

4 a^{2}

to the left side

2 a^{2} - 2 a - 3 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 2 (- 3) = 27

a_{1, 2} = \frac{- ( - 2 ) \pm 2 7}{2 \cdot 2}

Separate the solutions

a_{1} = \frac{- ( - 2 ) + 2 7}{2 \cdot 2}, a_{2} = \frac{- ( - 2 ) - 2 7}{2 \cdot 2}

a = \frac{- ( - 2 ) + 2 7}{2 \cdot 2} : \frac{1 + 7}{2}

a = \frac{- ( - 2 ) - 2 7}{2 \cdot 2} : \frac{1 - 7}{2}

The solutions to the quadratic equation are:

a = \frac{1 + 7}{2}, a = \frac{1 - 7}{2}

\frac{\frac{1}{x}}{\frac{1}{y}}

6^{2} + 1 1^{2} = x^{2}

k^{2} + 5 k - 14

n^{2} - 2 n + 1

27 m^{3} + 64 n^{9}

What are the solutions to the equation -2a^2-2a-3=-4a^2 ?
The solutions to the equation -2a^2-2a-3=-4a^2 are a=(1+sqrt(7))/2 ,a=(1-sqrt(7))/2
Find the zeros of -2a^2-2a-3=-4a^2
The zeros of -2a^2-2a-3=-4a^2 are a=(1+sqrt(7))/2 ,a=(1-sqrt(7))/2