What are the solutions to the equation a^2-7a+12=0 ?

The solutions to the equation a^2-7a+12=0 are a=4,a=3

Find the zeros of a^2-7a+12=0

The zeros of a^2-7a+12=0 are a=4,a=3

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a^{2} - 7 a + 12 = 0

a = 4, a = 3

Solution steps

a^{2} - 7 a + 12 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot 12 = 1

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

Separate the solutions

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

a = \frac{- ( - 7 ) + 1}{2 \cdot 1} : 4

a = \frac{- ( - 7 ) - 1}{2 \cdot 1} : 3

The solutions to the quadratic equation are:

a = 4, a = 3

3 [- 2 (4 - X)] - 5 = 2 X - 5

- 2 t^{2} + 10 t = 0

3 (- 2 x + 4) - 4 x - 2 = - 20

3^{2} + x^{2} = 7^{2}

\frac{3}{4} \cdot \frac{1}{5}

What are the solutions to the equation a^2-7a+12=0 ?
The solutions to the equation a^2-7a+12=0 are a=4,a=3
Find the zeros of a^2-7a+12=0
The zeros of a^2-7a+12=0 are a=4,a=3