What are the solutions to the equation 3x^2+ax=a^2 ?

The solutions to the equation 3x^2+ax=a^2 are x=(-a+sqrt(13)a)/6 ,x=(-a-sqrt(13)a)/6

Find the zeros of 3x^2+ax=a^2

The zeros of 3x^2+ax=a^2 are x=(-a+sqrt(13)a)/6 ,x=(-a-sqrt(13)a)/6

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3 x^{2} + a x = a^{2}

x = \frac{- a + 13 a}{6}, x = \frac{- a - 13 a}{6}

Solution steps

3 x^{2} + a x = a^{2}

Move

a^{2}

to the left side

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

Solve with the quadratic formula

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

Simplify

a^{2} - 4 \cdot 3 (- a^{2}) : 13 a

x_{1, 2} = \frac{- a \pm 13 a}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- a + 13 a}{2 \cdot 3}, x_{2} = \frac{- a - 13 a}{2 \cdot 3}

x = \frac{- a + 13 a}{2 \cdot 3} : \frac{- a + 13 a}{6}

x = \frac{- a - 13 a}{2 \cdot 3} : \frac{- a - 13 a}{6}

The solutions to the quadratic equation are:

x = \frac{- a + 13 a}{6}, x = \frac{- a - 13 a}{6}

7 x^{2} - 3 = - 11

4 (x^{2} - 5) + 5 = - 3

14 x^{2} - 108 x + 144 = 0

4 x^{2} + 17 = 20 x

y, x^{2} + 4 x y = 16

What are the solutions to the equation 3x^2+ax=a^2 ?
The solutions to the equation 3x^2+ax=a^2 are x=(-a+sqrt(13)a)/6 ,x=(-a-sqrt(13)a)/6
Find the zeros of 3x^2+ax=a^2
The zeros of 3x^2+ax=a^2 are x=(-a+sqrt(13)a)/6 ,x=(-a-sqrt(13)a)/6