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

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

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

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

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

b = 4, b = 3

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

b = 4, b = 3

6 x^{2} - 15 = 2 x^{2} + 10

2 X^{2} - 12 X = 0

5 x = 4 x^{2} - 2

y, (y + 1)^{2} = (2 - x) ln (x)

4 x^{2} - 6 x + 9

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