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

The solutions to the equation a^2-4=3a are a=4,a=-1

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

The zeros of a^2-4=3a are a=4,a=-1

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a^{2} - 4 = 3 a

a = 4, a = - 1

Solution steps

a^{2} - 4 = 3 a

Move

3 a

to the left side

a^{2} - 4 - 3 a = 0

Write in the standard form

a x^{2} + b x + c = 0

a^{2} - 3 a - 4 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 4) = 5

a_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

a_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, a_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

a = \frac{- ( - 3 ) + 5}{2 \cdot 1} : 4

a = \frac{- ( - 3 ) - 5}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

a = 4, a = - 1

4 x^{2} + 27 x - 18 = 0

\frac{6 x ^{2} + 2 x}{3} = 20 x

x^{2} - 9 x = - 29

x^{2} - 6 x + 9 = - 3

- 9 x^{2} + 153 x - 144 = 0

What are the solutions to the equation a^2-4=3a ?
The solutions to the equation a^2-4=3a are a=4,a=-1
Find the zeros of a^2-4=3a
The zeros of a^2-4=3a are a=4,a=-1