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

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

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

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

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

s = 4, s = 3

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

s = 4, s = 3

5 x + 11 < 26

53 \cdot 23

625 m^{4} - 81

2 z + 1 = z

∣ y ∣ > 5

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