What are the solutions to the equation pi=q(20-1q-1) ?

The solutions to the equation pi=q(20-1q-1) are q=(-sqrt(-4pi+361)+19)/2 ,q=(sqrt(-4pi+361)+19)/2

Find the zeros of pi=q(20-1q-1)

The zeros of pi=q(20-1q-1) are q=(-sqrt(-4pi+361)+19)/2 ,q=(sqrt(-4pi+361)+19)/2

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π = q (20 - 1 q - 1)

q = \frac{- - 4 π + 361 + 19}{2}, q = \frac{- 4 π + 361 + 19}{2}

Decimal Notation

q = 0.16681 \dots, q = 18.83318 \dots

Solution steps

π = q (20 - 1 \cdot q - 1)

Expand

q (20 - 1 \cdot q - 1) : - q^{2} + 19 q

π = - q^{2} + 19 q

Switch sides

- q^{2} + 19 q = π

Move

π

to the left side

- q^{2} + 19 q - π = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 ( - 1 ) ( - π )}{2 ( - 1 )}

1 9^{2} - 4 (- 1) (- π) = 361 - 4 π

q_{1, 2} = \frac{- 19 \pm 361 - 4 π}{2 ( - 1 )}

Separate the solutions

q_{1} = \frac{- 19 + 361 - 4 π}{2 ( - 1 )}, q_{2} = \frac{- 19 - 361 - 4 π}{2 ( - 1 )}

q = \frac{- 19 + 361 - 4 π}{2 ( - 1 )} : \frac{- - 4 π + 361 + 19}{2}

q = \frac{- 19 - 361 - 4 π}{2 ( - 1 )} : \frac{- 4 π + 361 + 19}{2}

The solutions to the quadratic equation are:

q = \frac{- - 4 π + 361 + 19}{2}, q = \frac{- 4 π + 361 + 19}{2}

4 w^{2} = - 12 w - 9

x^{2} + 9 x = 8

11.11 x = 1.1 x^{2}

(x^{2} - x + 3) = 5

x, f (x - 1) = x^{2} + 4 x - 5

What are the solutions to the equation pi=q(20-1q-1) ?
The solutions to the equation pi=q(20-1q-1) are q=(-sqrt(-4pi+361)+19)/2 ,q=(sqrt(-4pi+361)+19)/2
Find the zeros of pi=q(20-1q-1)
The zeros of pi=q(20-1q-1) are q=(-sqrt(-4pi+361)+19)/2 ,q=(sqrt(-4pi+361)+19)/2