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

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

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

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

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

y = 4, y = 3

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 4, y = 3

42 = 6 x

lo g_{2} (2 x - x^{2}) < 0

lo g_{4} (8) + lo g_{4} (2)

- 3 x^{4} - 7 x^{3} - 3 x^{2} + 3 x + 2

3 a^{2} = 21 a

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