What are the solutions to the equation (q)=0.1q^2+3q+2 ?

The solutions to the equation (q)=0.1q^2+3q+2 are q=2(2sqrt(5)-5),q=-2(5+2sqrt(5))

Find the zeros of (q)=0.1q^2+3q+2

The zeros of (q)=0.1q^2+3q+2 are q=2(2sqrt(5)-5),q=-2(5+2sqrt(5))

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(q) = 0.1 q^{2} + 3 q + 2

q = 2 (25 - 5), q = - 2 (5 + 25)

Decimal Notation

q = - 1.05572 \dots, q = - 18.94427 \dots

Solution steps

(q) = 0.1 q^{2} + 3 q + 2

Multiply both sides by

10

q \cdot 10 = q^{2} + 30 q + 20

Switch sides

q^{2} + 30 q + 20 = q \cdot 10

Move

q 10

to the left side

q^{2} + 20 q + 20 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

2 0^{2} - 4 \cdot 1 \cdot 20 = 85

q_{1, 2} = \frac{- 20 \pm 8 5}{2 \cdot 1}

Separate the solutions

q_{1} = \frac{- 20 + 8 5}{2 \cdot 1}, q_{2} = \frac{- 20 - 8 5}{2 \cdot 1}

q = \frac{- 20 + 8 5}{2 \cdot 1} : 2 (25 - 5)

q = \frac{- 20 - 8 5}{2 \cdot 1} : - 2 (5 + 25)

The solutions to the quadratic equation are:

q = 2 (25 - 5), q = - 2 (5 + 25)

8 t^{2} + 15 t + 15 = 0

7^{2} = 1^{2} + x^{2}

x (x + 7) = 5 x^{2} + 4

- 3 x^{2} = 21 x

(y - 7)^{2} - 8 = 0

What are the solutions to the equation (q)=0.1q^2+3q+2 ?
The solutions to the equation (q)=0.1q^2+3q+2 are q=2(2sqrt(5)-5),q=-2(5+2sqrt(5))
Find the zeros of (q)=0.1q^2+3q+2
The zeros of (q)=0.1q^2+3q+2 are q=2(2sqrt(5)-5),q=-2(5+2sqrt(5))