What are the solutions to the equation 2(s+1)^2-5(s+1)-3=0 ?

The solutions to the equation 2(s+1)^2-5(s+1)-3=0 are s=2,s=-3/2

Find the zeros of 2(s+1)^2-5(s+1)-3=0

The zeros of 2(s+1)^2-5(s+1)-3=0 are s=2,s=-3/2

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2 (s + 1)^{2} - 5 (s + 1) - 3 = 0

s = 2, s = - \frac{3}{2}

Decimal Notation

s = 2, s = - 1.5

Solution steps

2 (s + 1)^{2} - 5 (s + 1) - 3 = 0

Expand

2 (s + 1)^{2} - 5 (s + 1) - 3 : 2 s^{2} - s - 6

2 s^{2} - s - 6 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 2 (- 6) = 7

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

s = 2, s = - \frac{3}{2}

x^{2} = 8^{2} + (- 15)^{2}

(- 2 - x) (1 - x) + 10 = 0

x^{2} + 85 = 20 x

x^{2} + 3 x - 1 = 4 x + 2

3 (3 x + 7) + (3 x + 7)^{2} = 0

What are the solutions to the equation 2(s+1)^2-5(s+1)-3=0 ?
The solutions to the equation 2(s+1)^2-5(s+1)-3=0 are s=2,s=-3/2
Find the zeros of 2(s+1)^2-5(s+1)-3=0
The zeros of 2(s+1)^2-5(s+1)-3=0 are s=2,s=-3/2