What are the solutions to the equation v^2-2v=23 ?

The solutions to the equation v^2-2v=23 are v=1+2sqrt(6),v=1-2sqrt(6)

Find the zeros of v^2-2v=23

The zeros of v^2-2v=23 are v=1+2sqrt(6),v=1-2sqrt(6)

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v^{2} - 2 v = 23

v = 1 + 26, v = 1 - 26

Decimal Notation

v = 5.89897 \dots, v = - 3.89897 \dots

Solution steps

v^{2} - 2 v = 23

Move

23

to the left side

v^{2} - 2 v - 23 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 23 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 23) = 46

v_{1, 2} = \frac{- ( - 2 ) \pm 4 6}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 2 ) + 4 6}{2 \cdot 1}, v_{2} = \frac{- ( - 2 ) - 4 6}{2 \cdot 1}

v = \frac{- ( - 2 ) + 4 6}{2 \cdot 1} : 1 + 26

v = \frac{- ( - 2 ) - 4 6}{2 \cdot 1} : 1 - 26

The solutions to the quadratic equation are:

v = 1 + 26, v = 1 - 26

12 = 2 x

w^{2} + (w + 3)^{2} = 1 3^{2}

x - 1 < 7

3 x^{2} = 4 x + 3

\frac{13}{7} \cdot 5

What are the solutions to the equation v^2-2v=23 ?
The solutions to the equation v^2-2v=23 are v=1+2sqrt(6),v=1-2sqrt(6)
Find the zeros of v^2-2v=23
The zeros of v^2-2v=23 are v=1+2sqrt(6),v=1-2sqrt(6)