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

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

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

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

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

r = 4, r = 3

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

r = 4, r = 3

π (4)^{2}

x^{2} = 36

- 8 \leq 4 (y + 10)

\frac{1}{x + y} - \frac{1}{x - y} + \frac{2 y}{x ^{2} + y ^{2}}

\frac{x - 2}{2 x + 9} \geq 1

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