What are the solutions to the equation 2d^2+5d+4=1 ?

The solutions to the equation 2d^2+5d+4=1 are d=-1,d=-3/2

Find the zeros of 2d^2+5d+4=1

The zeros of 2d^2+5d+4=1 are d=-1,d=-3/2

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2 d^{2} + 5 d + 4 = 1

d = - 1, d = - \frac{3}{2}

Decimal Notation

d = - 1, d = - 1.5

Solution steps

2 d^{2} + 5 d + 4 = 1

Move

1

to the left side

2 d^{2} + 5 d + 3 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

5^{2} - 4 \cdot 2 \cdot 3 = 1

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

d = - 1, d = - \frac{3}{2}

x^{2} + (m + 1) x + 16 = 0

λ^{2} - 2 λ - 3 = 0

λ^{2} + aλ - 2 a^{2} = 0

(18 - x)^{2} + (5 - x)^{2} = 81

m^{2} + 3 = 51

What are the solutions to the equation 2d^2+5d+4=1 ?
The solutions to the equation 2d^2+5d+4=1 are d=-1,d=-3/2
Find the zeros of 2d^2+5d+4=1
The zeros of 2d^2+5d+4=1 are d=-1,d=-3/2