What are the solutions to the equation 2d^2-19d+14=-6d ?

The solutions to the equation 2d^2-19d+14=-6d are d=(13+sqrt(57))/4 ,d=(13-sqrt(57))/4

Find the zeros of 2d^2-19d+14=-6d

The zeros of 2d^2-19d+14=-6d are d=(13+sqrt(57))/4 ,d=(13-sqrt(57))/4

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2 d^{2} - 19 d + 14 = - 6 d

d = \frac{13 + 57}{4}, d = \frac{13 - 57}{4}

Decimal Notation

d = 5.13745 \dots, d = 1.36254 \dots

Solution steps

2 d^{2} - 19 d + 14 = - 6 d

Move

6 d

to the left side

2 d^{2} - 13 d + 14 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 2 \cdot 14}{2 \cdot 2}

(- 13)^{2} - 4 \cdot 2 \cdot 14 = 57

d_{1, 2} = \frac{- ( - 13 ) \pm 57}{2 \cdot 2}

Separate the solutions

d_{1} = \frac{- ( - 13 ) + 57}{2 \cdot 2}, d_{2} = \frac{- ( - 13 ) - 57}{2 \cdot 2}

d = \frac{- ( - 13 ) + 57}{2 \cdot 2} : \frac{13 + 57}{4}

d = \frac{- ( - 13 ) - 57}{2 \cdot 2} : \frac{13 - 57}{4}

The solutions to the quadratic equation are:

d = \frac{13 + 57}{4}, d = \frac{13 - 57}{4}

\frac{1}{6} + \frac{1}{3}

(16 x^{2} y^{8})^{- \frac{1}{2}}

(3 x + 2) \geq 5

x + 42 = 5 x + 10

- 12 x = 4

What are the solutions to the equation 2d^2-19d+14=-6d ?
The solutions to the equation 2d^2-19d+14=-6d are d=(13+sqrt(57))/4 ,d=(13-sqrt(57))/4
Find the zeros of 2d^2-19d+14=-6d
The zeros of 2d^2-19d+14=-6d are d=(13+sqrt(57))/4 ,d=(13-sqrt(57))/4