What are the solutions to the equation 16v^2-40v+31=6 ?

The solutions to the equation 16v^2-40v+31=6 are v= 5/4

Find the zeros of 16v^2-40v+31=6

The zeros of 16v^2-40v+31=6 are v= 5/4

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16 v^{2} - 40 v + 31 = 6

v = \frac{5}{4}

Decimal Notation

v = 1.25

Solution steps

16 v^{2} - 40 v + 31 = 6

Move

6

to the left side

16 v^{2} - 40 v + 25 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 \cdot 16 \cdot 25}{2 \cdot 16}

(- 40)^{2} - 4 \cdot 16 \cdot 25 = 0

v_{1, 2} = \frac{- ( - 40 ) \pm 0}{2 \cdot 16}

v = \frac{- ( - 40 )}{2 \cdot 16}

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

v = \frac{5}{4}

The solution to the quadratic equation is:

v = \frac{5}{4}

5 x^{2} + 4 x = - 4

x^{2} + 7 x + 10

x^{2} + 2 x = 0

λ^{2} + λ - 6 (1) = 0

(2 s^{2} + 8 s + 16) = 0

What are the solutions to the equation 16v^2-40v+31=6 ?
The solutions to the equation 16v^2-40v+31=6 are v= 5/4
Find the zeros of 16v^2-40v+31=6
The zeros of 16v^2-40v+31=6 are v= 5/4