What are the solutions to the equation 4v^2+23v+16=1 ?

The solutions to the equation 4v^2+23v+16=1 are v=-3/4 ,v=-5

Find the zeros of 4v^2+23v+16=1

The zeros of 4v^2+23v+16=1 are v=-3/4 ,v=-5

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4 v^{2} + 23 v + 16 = 1

v = - \frac{3}{4}, v = - 5

Decimal Notation

v = - 0.75, v = - 5

Solution steps

4 v^{2} + 23 v + 16 = 1

Move

1

to the left side

4 v^{2} + 23 v + 15 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- 23 \pm 2 3 ^{2} - 4 \cdot 4 \cdot 15}{2 \cdot 4}

2 3^{2} - 4 \cdot 4 \cdot 15 = 17

v_{1, 2} = \frac{- 23 \pm 17}{2 \cdot 4}

Separate the solutions

v_{1} = \frac{- 23 + 17}{2 \cdot 4}, v_{2} = \frac{- 23 - 17}{2 \cdot 4}

v = \frac{- 23 + 17}{2 \cdot 4} : - \frac{3}{4}

v = \frac{- 23 - 17}{2 \cdot 4} : - 5

The solutions to the quadratic equation are:

v = - \frac{3}{4}, v = - 5

x^{2} + 12 x = 72

2 k^{2} + 15 k + 19 = 0

2 x^{2} - 1 = 3 x + 4

x^{2} - \frac{10}{3} x + 1 = 0

y^{2} + 16 y + 13 = 0

What are the solutions to the equation 4v^2+23v+16=1 ?
The solutions to the equation 4v^2+23v+16=1 are v=-3/4 ,v=-5
Find the zeros of 4v^2+23v+16=1
The zeros of 4v^2+23v+16=1 are v=-3/4 ,v=-5