What are the solutions to the equation 4v^2-15v+16=6 ?

The solutions to the equation 4v^2-15v+16=6 are v=(15+sqrt(65))/8 ,v=(15-sqrt(65))/8

Find the zeros of 4v^2-15v+16=6

The zeros of 4v^2-15v+16=6 are v=(15+sqrt(65))/8 ,v=(15-sqrt(65))/8

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4 v^{2} - 15 v + 16 = 6

v = \frac{15 + 65}{8}, v = \frac{15 - 65}{8}

Decimal Notation

v = 2.88278 \dots, v = 0.86721 \dots

Solution steps

4 v^{2} - 15 v + 16 = 6

Move

6

to the left side

4 v^{2} - 15 v + 10 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 4 \cdot 10}{2 \cdot 4}

(- 15)^{2} - 4 \cdot 4 \cdot 10 = 65

v_{1, 2} = \frac{- ( - 15 ) \pm 65}{2 \cdot 4}

Separate the solutions

v_{1} = \frac{- ( - 15 ) + 65}{2 \cdot 4}, v_{2} = \frac{- ( - 15 ) - 65}{2 \cdot 4}

v = \frac{- ( - 15 ) + 65}{2 \cdot 4} : \frac{15 + 65}{8}

v = \frac{- ( - 15 ) - 65}{2 \cdot 4} : \frac{15 - 65}{8}

The solutions to the quadratic equation are:

v = \frac{15 + 65}{8}, v = \frac{15 - 65}{8}

- 3 x + 6 < 15 or 4 x - 1 \geq 3

\frac{3}{4} x = \frac{1}{2}

- 1 + 4 x \leq 31

\frac{3 x}{2} \leq 9

10 x^{5} y - 40 x y^{3}

What are the solutions to the equation 4v^2-15v+16=6 ?
The solutions to the equation 4v^2-15v+16=6 are v=(15+sqrt(65))/8 ,v=(15-sqrt(65))/8
Find the zeros of 4v^2-15v+16=6
The zeros of 4v^2-15v+16=6 are v=(15+sqrt(65))/8 ,v=(15-sqrt(65))/8