What are the solutions to the equation 2d^2+15d+23=3 ?

The solutions to the equation 2d^2+15d+23=3 are d=(-15+sqrt(65))/4 ,d=(-15-sqrt(65))/4

Find the zeros of 2d^2+15d+23=3

The zeros of 2d^2+15d+23=3 are d=(-15+sqrt(65))/4 ,d=(-15-sqrt(65))/4

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2 d^{2} + 15 d + 23 = 3

d = \frac{- 15 + 65}{4}, d = \frac{- 15 - 65}{4}

Decimal Notation

d = - 1.73443 \dots, d = - 5.76556 \dots

Solution steps

2 d^{2} + 15 d + 23 = 3

Move

3

to the left side

2 d^{2} + 15 d + 20 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 2 \cdot 20}{2 \cdot 2}

1 5^{2} - 4 \cdot 2 \cdot 20 = 65

d_{1, 2} = \frac{- 15 \pm 65}{2 \cdot 2}

Separate the solutions

d_{1} = \frac{- 15 + 65}{2 \cdot 2}, d_{2} = \frac{- 15 - 65}{2 \cdot 2}

d = \frac{- 15 + 65}{2 \cdot 2} : \frac{- 15 + 65}{4}

d = \frac{- 15 - 65}{2 \cdot 2} : \frac{- 15 - 65}{4}

The solutions to the quadratic equation are:

d = \frac{- 15 + 65}{4}, d = \frac{- 15 - 65}{4}

3 n^{3} - 3 n

n^{\frac{1}{2 ^{l o g_{2} (l o g_{2} (n))}}}

20 (\frac{π}{4})

\frac{3}{4} (x - 1) - \frac{1}{2} (3 x + 2) = 0

\frac{2 ^{n + 1}}{2 ^{n}}

What are the solutions to the equation 2d^2+15d+23=3 ?
The solutions to the equation 2d^2+15d+23=3 are d=(-15+sqrt(65))/4 ,d=(-15-sqrt(65))/4
Find the zeros of 2d^2+15d+23=3
The zeros of 2d^2+15d+23=3 are d=(-15+sqrt(65))/4 ,d=(-15-sqrt(65))/4