What are the solutions to the equation d^2+d-7=-5 ?

The solutions to the equation d^2+d-7=-5 are d=1,d=-2

Find the zeros of d^2+d-7=-5

The zeros of d^2+d-7=-5 are d=1,d=-2

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d^{2} + d - 7 = - 5

d = 1, d = - 2

Solution steps

d^{2} + d - 7 = - 5

Move

5

to the left side

d^{2} + d - 2 = 0

Solve with the quadratic formula

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

1^{2} - 4 \cdot 1 \cdot (- 2) = 3

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

Separate the solutions

d_{1} = \frac{- 1 + 3}{2 \cdot 1}, d_{2} = \frac{- 1 - 3}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

d = 1, d = - 2

\frac{1}{2} (2)

8 x - x^{2} = 15

1 5^{\frac{3}{2}}

166 = - 2 (- 5 + 2 x) - 9 x

- \frac{2}{9} f = 12

What are the solutions to the equation d^2+d-7=-5 ?
The solutions to the equation d^2+d-7=-5 are d=1,d=-2
Find the zeros of d^2+d-7=-5
The zeros of d^2+d-7=-5 are d=1,d=-2