What are the solutions to the equation (q)=0.4q^2-40q+40 ?

The solutions to the equation (q)=0.4q^2-40q+40 are q=(5(41+7sqrt(33)))/4 ,q=(5(41-7sqrt(33)))/4

Find the zeros of (q)=0.4q^2-40q+40

The zeros of (q)=0.4q^2-40q+40 are q=(5(41+7sqrt(33)))/4 ,q=(5(41-7sqrt(33)))/4

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(q) = 0.4 q^{2} - 40 q + 40

q = \frac{5 ( 41 + 7 33 )}{4}, q = \frac{5 ( 41 - 7 33 )}{4}

Decimal Notation

q = 101.51492 \dots, q = 0.98507 \dots

Solution steps

(q) = 0.4 q^{2} - 40 q + 40

Multiply both sides by

10

q \cdot 10 = 4 q^{2} - 400 q + 400

Switch sides

4 q^{2} - 400 q + 400 = q \cdot 10

Move

q 10

to the left side

4 q^{2} - 410 q + 400 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 410 ) \pm ( - 410 ) ^{2} - 4 \cdot 4 \cdot 400}{2 \cdot 4}

(- 410)^{2} - 4 \cdot 4 \cdot 400 = 7033

q_{1, 2} = \frac{- ( - 410 ) \pm 70 33}{2 \cdot 4}

Separate the solutions

q_{1} = \frac{- ( - 410 ) + 70 33}{2 \cdot 4}, q_{2} = \frac{- ( - 410 ) - 70 33}{2 \cdot 4}

q = \frac{- ( - 410 ) + 70 33}{2 \cdot 4} : \frac{5 ( 41 + 7 33 )}{4}

q = \frac{- ( - 410 ) - 70 33}{2 \cdot 4} : \frac{5 ( 41 - 7 33 )}{4}

The solutions to the quadratic equation are:

q = \frac{5 ( 41 + 7 33 )}{4}, q = \frac{5 ( 41 - 7 33 )}{4}

x^{2} + 4 x + 1 = 7 - x^{2}

2 x^{2} + 3 x - 4 = 0

- 3 x^{2} = 72

2 x^{2} + 4 x - 3 = 0

4.9 t^{2} - 25 t - 50 = 0

What are the solutions to the equation (q)=0.4q^2-40q+40 ?
The solutions to the equation (q)=0.4q^2-40q+40 are q=(5(41+7sqrt(33)))/4 ,q=(5(41-7sqrt(33)))/4
Find the zeros of (q)=0.4q^2-40q+40
The zeros of (q)=0.4q^2-40q+40 are q=(5(41+7sqrt(33)))/4 ,q=(5(41-7sqrt(33)))/4