Solve 5(100 - 4k)k = 2

Question

Solve the quadratic equation

Solve using the quadratic formula

Solve by completing the square

Solve using the PQ formula

k_{1} = \frac{125 - 3 1735}{10}, k_{2} = \frac{125 + 3 1735}{10}

Alternative Form

k_{1} \approx 0.004001, k_{2} \approx 24.995999

Evaluate

5 (100 - 4 k) k = 2

Multiply the terms

5 k (100 - 4 k) = 2

Expand the expression

More Steps

500 k - 20 k^{2} = 2

Move the expression to the left side

500 k - 20 k^{2} - 2 = 0

Rewrite in standard form

- 20 k^{2} + 500 k - 2 = 0

Multiply both sides

20 k^{2} - 500 k + 2 = 0

Substitute a= 20,b= - 500 and c= 2 into the quadratic formula k = \frac{- b \pm b ^{2} - 4 a c}{2 a}

k = \frac{500 \pm ( - 500 ) ^{2} - 4 \times 20 \times 2}{2 \times 20}

Simplify the expression

k = \frac{500 \pm ( - 500 ) ^{2} - 4 \times 20 \times 2}{40}

Simplify the expression

More Steps

k = \frac{500 \pm 50 0 ^{2} - 160}{40}

Simplify the radical expression

More Steps

k = \frac{500 \pm 12 1735}{40}

Separate the equation into 2 possible cases

k = \frac{500 + 12 1735}{40} k = \frac{500 - 12 1735}{40}

Simplify the expression

More Steps

k = \frac{125 + 3 1735}{10} k = \frac{500 - 12 1735}{40}

Simplify the expression

More Steps

k = \frac{125 + 3 1735}{10} k = \frac{125 - 3 1735}{10}

Solution

k_{1} = \frac{125 - 3 1735}{10}, k_{2} = \frac{125 + 3 1735}{10}

Alternative Form

k_{1} \approx 0.004001, k_{2} \approx 24.995999

Show Solution