What are the solutions to the equation 2400(d)=-d^2+70d+1800 ?

The solutions to the equation 2400(d)=-d^2+70d+1800 are d=-5(233+sqrt(54361)),d=5(sqrt(54361)-233)

Find the zeros of 2400(d)=-d^2+70d+1800

The zeros of 2400(d)=-d^2+70d+1800 are d=-5(233+sqrt(54361)),d=5(sqrt(54361)-233)

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2400 (d) = - d^{2} + 70 d + 1800

d = - 5 (233 + 54361), d = 5 (54361 - 233)

Decimal Notation

d = - 2330.77227 \dots, d = 0.77227 \dots

Solution steps

2400 (d) = - d^{2} + 70 d + 1800

Refine

2400 d = - d^{2} + 70 d + 1800

Switch sides

- d^{2} + 70 d + 1800 = 2400 d

Move

2400 d

to the left side

- d^{2} - 2330 d + 1800 = 0

Solve with the quadratic formula

d_{1, 2} = \frac{- ( - 2330 ) \pm ( - 2330 ) ^{2} - 4 ( - 1 ) \cdot 1800}{2 ( - 1 )}

(- 2330)^{2} - 4 (- 1) \cdot 1800 = 1054361

d_{1, 2} = \frac{- ( - 2330 ) \pm 10 54361}{2 ( - 1 )}

Separate the solutions

d_{1} = \frac{- ( - 2330 ) + 10 54361}{2 ( - 1 )}, d_{2} = \frac{- ( - 2330 ) - 10 54361}{2 ( - 1 )}

d = \frac{- ( - 2330 ) + 10 54361}{2 ( - 1 )} : - 5 (233 + 54361)

d = \frac{- ( - 2330 ) - 10 54361}{2 ( - 1 )} : 5 (54361 - 233)

The solutions to the quadratic equation are:

d = - 5 (233 + 54361), d = 5 (54361 - 233)

x^{2} - 12 x - 17 = 0

- 49 x + x^{2} = 0

k^{2} + 10 k - 3 = 0

x, 2 x^{2} - 5 x + 3 = 0

x^{2} - 8 = 12

What are the solutions to the equation 2400(d)=-d^2+70d+1800 ?
The solutions to the equation 2400(d)=-d^2+70d+1800 are d=-5(233+sqrt(54361)),d=5(sqrt(54361)-233)
Find the zeros of 2400(d)=-d^2+70d+1800
The zeros of 2400(d)=-d^2+70d+1800 are d=-5(233+sqrt(54361)),d=5(sqrt(54361)-233)