What are the solutions to the equation -a^2-3a-13=-2a^2 ?

The solutions to the equation -a^2-3a-13=-2a^2 are a=(3+sqrt(61))/2 ,a=(3-sqrt(61))/2

Find the zeros of -a^2-3a-13=-2a^2

The zeros of -a^2-3a-13=-2a^2 are a=(3+sqrt(61))/2 ,a=(3-sqrt(61))/2

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- a^{2} - 3 a - 13 = - 2 a^{2}

a = \frac{3 + 61}{2}, a = \frac{3 - 61}{2}

Decimal Notation

a = 5.40512 \dots, a = - 2.40512 \dots

Solution steps

- a^{2} - 3 a - 13 = - 2 a^{2}

Move

2 a^{2}

to the left side

a^{2} - 3 a - 13 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 13 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 13) = 61

a_{1, 2} = \frac{- ( - 3 ) \pm 61}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- ( - 3 ) + 61}{2 \cdot 1}, a_{2} = \frac{- ( - 3 ) - 61}{2 \cdot 1}

a = \frac{- ( - 3 ) + 61}{2 \cdot 1} : \frac{3 + 61}{2}

a = \frac{- ( - 3 ) - 61}{2 \cdot 1} : \frac{3 - 61}{2}

The solutions to the quadratic equation are:

a = \frac{3 + 61}{2}, a = \frac{3 - 61}{2}

y^{2} + 9 y + 3 = - 4 y^{2}

\frac{1}{4} (x + 1)^{2} - 16 = 0

2 x^{2} - 3 x = 3

x, \frac{x ^{2} + y ^{2}}{z ^{2}} = - w^{2}

2 x^{2} - 14 x = - 9 x + 7

What are the solutions to the equation -a^2-3a-13=-2a^2 ?
The solutions to the equation -a^2-3a-13=-2a^2 are a=(3+sqrt(61))/2 ,a=(3-sqrt(61))/2
Find the zeros of -a^2-3a-13=-2a^2
The zeros of -a^2-3a-13=-2a^2 are a=(3+sqrt(61))/2 ,a=(3-sqrt(61))/2