What are the solutions to the equation k(k+40)-10k+20k=490 ?

The solutions to the equation k(k+40)-10k+20k=490 are k=-25+sqrt(1115),k=-25-sqrt(1115)

Find the zeros of k(k+40)-10k+20k=490

The zeros of k(k+40)-10k+20k=490 are k=-25+sqrt(1115),k=-25-sqrt(1115)

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k (k + 40) - 10 k + 20 k = 490

k = - 25 + 1115, k = - 25 - 1115

Decimal Notation

k = 8.39161 \dots, k = - 58.39161 \dots

Solution steps

k (k + 40) - 10 k + 20 k = 490

Expand

k (k + 40) - 10 k + 20 k : k^{2} + 50 k

k^{2} + 50 k = 490

Move

490

to the left side

k^{2} + 50 k - 490 = 0

Solve with the quadratic formula

k_{1, 2} = \frac{- 50 \pm 5 0 ^{2} - 4 \cdot 1 \cdot ( - 490 )}{2 \cdot 1}

5 0^{2} - 4 \cdot 1 \cdot (- 490) = 21115

k_{1, 2} = \frac{- 50 \pm 2 1115}{2 \cdot 1}

Separate the solutions

k_{1} = \frac{- 50 + 2 1115}{2 \cdot 1}, k_{2} = \frac{- 50 - 2 1115}{2 \cdot 1}

k = \frac{- 50 + 2 1115}{2 \cdot 1} : - 25 + 1115

k = \frac{- 50 - 2 1115}{2 \cdot 1} : - 25 - 1115

The solutions to the quadratic equation are:

k = - 25 + 1115, k = - 25 - 1115

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r^{2} + 10 r + 29 = 0

\frac{3 x}{2} - 10 = \frac{1}{2}

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

x^{2} - 8 x - 13 = 0

What are the solutions to the equation k(k+40)-10k+20k=490 ?
The solutions to the equation k(k+40)-10k+20k=490 are k=-25+sqrt(1115),k=-25-sqrt(1115)
Find the zeros of k(k+40)-10k+20k=490
The zeros of k(k+40)-10k+20k=490 are k=-25+sqrt(1115),k=-25-sqrt(1115)