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

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

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

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

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a^{2} - 3 a = 4

a = 4, a = - 1

Solution steps

a^{2} - 3 a = 4

Move

4

to the left side

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

a^{2} - 3 a = 0

6 r^{2} + 9 r + 7 = 0

(x + 188) (x + 380) = 0

5 x^{2} - 120 = 5

\frac{1}{20} x^{2} + 4 x + 100 = 0

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