What are the solutions to the equation (d+1)(d-5)=2d^2-7 ?

The solutions to the equation (d+1)(d-5)=2d^2-7 are d=-2-sqrt(6),d=sqrt(6)-2

Find the zeros of (d+1)(d-5)=2d^2-7

The zeros of (d+1)(d-5)=2d^2-7 are d=-2-sqrt(6),d=sqrt(6)-2

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(d + 1) (d - 5) = 2 d^{2} - 7

d = - 2 - 6, d = 6 - 2

Decimal Notation

d = - 4.44948 \dots, d = 0.44948 \dots

Solution steps

(d + 1) (d - 5) = 2 d^{2} - 7

Expand

(d + 1) (d - 5) : d^{2} - 4 d - 5

d^{2} - 4 d - 5 = 2 d^{2} - 7

Move

7

to the left side

d^{2} - 4 d + 2 = 2 d^{2}

Move

2 d^{2}

to the left side

- d^{2} - 4 d + 2 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 (- 1) \cdot 2 = 26

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

Separate the solutions

d_{1} = \frac{- ( - 4 ) + 2 6}{2 ( - 1 )}, d_{2} = \frac{- ( - 4 ) - 2 6}{2 ( - 1 )}

d = \frac{- ( - 4 ) + 2 6}{2 ( - 1 )} : - 2 - 6

d = \frac{- ( - 4 ) - 2 6}{2 ( - 1 )} : 6 - 2

The solutions to the quadratic equation are:

d = - 2 - 6, d = 6 - 2

4 = 3 t^{2} - t

k^{2} + 15 k = - 56

(x + 5) (x - 1) - 7 (x + 1) = - 3 x - 3

p^{2} + 3 p + 2 = 0

x, x (x + 5) = - 6

What are the solutions to the equation (d+1)(d-5)=2d^2-7 ?
The solutions to the equation (d+1)(d-5)=2d^2-7 are d=-2-sqrt(6),d=sqrt(6)-2
Find the zeros of (d+1)(d-5)=2d^2-7
The zeros of (d+1)(d-5)=2d^2-7 are d=-2-sqrt(6),d=sqrt(6)-2