What are the solutions to the equation s/2+(s^2)/2-3=0 ?

The solutions to the equation s/2+(s^2)/2-3=0 are s=2,s=-3

Find the zeros of s/2+(s^2)/2-3=0

The zeros of s/2+(s^2)/2-3=0 are s=2,s=-3

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\frac{s}{2} + \frac{s ^{2}}{2} - 3 = 0

s = 2, s = - 3

Solution steps

\frac{s}{2} + \frac{s ^{2}}{2} - 3 = 0

Multiply both sides by

2

s^{2} + s - 6 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 6) = 5

s_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- 1 + 5}{2 \cdot 1}, s_{2} = \frac{- 1 - 5}{2 \cdot 1}

s = \frac{- 1 + 5}{2 \cdot 1} : 2

s = \frac{- 1 - 5}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

s = 2, s = - 3

5 a^{2} + 6 a - 147 = - 3

x^{2} = 4^{2} + 2^{2}

6 x^{2} + x - 3 = 0

8 x^{2} - 5 x + 3 = 0

x^{2} - 3 x + 2 = 0

What are the solutions to the equation s/2+(s^2)/2-3=0 ?
The solutions to the equation s/2+(s^2)/2-3=0 are s=2,s=-3
Find the zeros of s/2+(s^2)/2-3=0
The zeros of s/2+(s^2)/2-3=0 are s=2,s=-3