What are the solutions to the equation 16x^2+40x=-25 ?

The solutions to the equation 16x^2+40x=-25 are x=-5/4

Find the zeros of 16x^2+40x=-25

The zeros of 16x^2+40x=-25 are x=-5/4

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16 x^{2} + 40 x = - 25

x = - \frac{5}{4}

Decimal Notation

x = - 1.25

Solution steps

16 x^{2} + 40 x = - 25

Move

25

to the left side

16 x^{2} + 40 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 \cdot 16 \cdot 25}{2 \cdot 16}

4 0^{2} - 4 \cdot 16 \cdot 25 = 0

x_{1, 2} = \frac{- 40 \pm 0}{2 \cdot 16}

x = \frac{- 40}{2 \cdot 16}

\frac{- 40}{2 \cdot 16} = - \frac{5}{4}

x = - \frac{5}{4}

The solution to the quadratic equation is:

x = - \frac{5}{4}

(x - 2) (x - 5) (x - 6) \leq 0

- 1.6 t + 16 = 0

\frac{\frac{15}{6}}{- \frac{7}{2}}

(\frac{4 x}{3 y})^{2}

(\frac{4}{6})^{2}

What are the solutions to the equation 16x^2+40x=-25 ?
The solutions to the equation 16x^2+40x=-25 are x=-5/4
Find the zeros of 16x^2+40x=-25
The zeros of 16x^2+40x=-25 are x=-5/4