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

The solutions to the equation d^2+d=6 are d=2,d=-3

Find the zeros of d^2+d=6

The zeros of d^2+d=6 are d=2,d=-3

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d^{2} + d = 6

d = 2, d = - 3

Solution steps

d^{2} + d = 6

Move

6

to the left side

d^{2} + d - 6 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

d = 2, d = - 3

45 x^{2} - 25

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

20 n + 15

2 x - 2 < - 12 or 2 x + 3 > 7

∣ 4 x - 7 ∣ \geq 6

What are the solutions to the equation d^2+d=6 ?
The solutions to the equation d^2+d=6 are d=2,d=-3
Find the zeros of d^2+d=6
The zeros of d^2+d=6 are d=2,d=-3