Solve 4 (-4x-5) = 5(-4x^2)

Question

Solve the quadratic equation

Solve using the quadratic formula

Solve by completing the square

Solve using the PQ formula

x_{1} = \frac{2 - 29}{5}, x_{2} = \frac{2 + 29}{5}

Alternative Form

x_{1} \approx - 0.677033, x_{2} \approx 1.477033

Evaluate

4 (- 4 x - 5) = 5 (- 4 x^{2})

Multiply the numbers

More Steps

4 (- 4 x - 5) = - 20 x^{2}

Swap the sides

- 20 x^{2} = 4 (- 4 x - 5)

Expand the expression

More Steps

- 20 x^{2} = - 16 x - 20

Move the expression to the left side

- 20 x^{2} + 16 x + 20 = 0

Multiply both sides

20 x^{2} - 16 x - 20 = 0

Substitute a= 20,b= - 16 and c= - 20 into the quadratic formula x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

x = \frac{16 \pm ( - 16 ) ^{2} - 4 \times 20 ( - 20 )}{2 \times 20}

Simplify the expression

x = \frac{16 \pm ( - 16 ) ^{2} - 4 \times 20 ( - 20 )}{40}

Simplify the expression

More Steps

x = \frac{16 \pm 1856}{40}

Simplify the radical expression

More Steps

x = \frac{16 \pm 8 29}{40}

Separate the equation into 2 possible cases

x = \frac{16 + 8 29}{40} x = \frac{16 - 8 29}{40}

Simplify the expression

More Steps

x = \frac{2 + 29}{5} x = \frac{16 - 8 29}{40}

Simplify the expression

More Steps

x = \frac{2 + 29}{5} x = \frac{2 - 29}{5}

Solution

x_{1} = \frac{2 - 29}{5}, x_{2} = \frac{2 + 29}{5}

Alternative Form

x_{1} \approx - 0.677033, x_{2} \approx 1.477033

Show Solution