What are the solutions to the equation 2a^2+4a-6=-3a ?

The solutions to the equation 2a^2+4a-6=-3a are a=(-7+sqrt(97))/4 ,a=(-7-sqrt(97))/4

Find the zeros of 2a^2+4a-6=-3a

The zeros of 2a^2+4a-6=-3a are a=(-7+sqrt(97))/4 ,a=(-7-sqrt(97))/4

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2 a^{2} + 4 a - 6 = - 3 a

a = \frac{- 7 + 97}{4}, a = \frac{- 7 - 97}{4}

Decimal

a = 0.71221 \dots, a = - 4.21221 \dots

Solution steps

2 a^{2} + 4 a - 6 = - 3 a

Move

3 a

to the left side

2 a^{2} + 7 a - 6 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 6) = 97

a_{1, 2} = \frac{- 7 \pm 97}{2 \cdot 2}

Separate the solutions

a_{1} = \frac{- 7 + 97}{2 \cdot 2}, a_{2} = \frac{- 7 - 97}{2 \cdot 2}

a = \frac{- 7 + 97}{2 \cdot 2} : \frac{- 7 + 97}{4}

a = \frac{- 7 - 97}{2 \cdot 2} : \frac{- 7 - 97}{4}

The solutions to the quadratic equation are:

a = \frac{- 7 + 97}{4}, a = \frac{- 7 - 97}{4}

y, 2 x^{2} - x y + y^{2} = 16

y, x^{2} = \frac{1}{y} - 1

x^{2} + \frac{2}{3} x - \frac{1}{9} = 0

4 m^{2} - 15 m + 5 = - 3 m

2 x^{2} - 13 x + 3 = - 4 x

What are the solutions to the equation 2a^2+4a-6=-3a ?
The solutions to the equation 2a^2+4a-6=-3a are a=(-7+sqrt(97))/4 ,a=(-7-sqrt(97))/4
Find the zeros of 2a^2+4a-6=-3a
The zeros of 2a^2+4a-6=-3a are a=(-7+sqrt(97))/4 ,a=(-7-sqrt(97))/4