What are the solutions to the equation 2q^2-4q+4=3q ?

The solutions to the equation 2q^2-4q+4=3q are q=(7+sqrt(17))/4 ,q=(7-sqrt(17))/4

Find the zeros of 2q^2-4q+4=3q

The zeros of 2q^2-4q+4=3q are q=(7+sqrt(17))/4 ,q=(7-sqrt(17))/4

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2 q^{2} - 4 q + 4 = 3 q

q = \frac{7 + 17}{4}, q = \frac{7 - 17}{4}

Decimal Notation

q = 2.78077 \dots, q = 0.71922 \dots

Solution steps

2 q^{2} - 4 q + 4 = 3 q

Move

3 q

to the left side

2 q^{2} - 7 q + 4 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 4}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 4 = 17

q_{1, 2} = \frac{- ( - 7 ) \pm 17}{2 \cdot 2}

Separate the solutions

q_{1} = \frac{- ( - 7 ) + 17}{2 \cdot 2}, q_{2} = \frac{- ( - 7 ) - 17}{2 \cdot 2}

q = \frac{- ( - 7 ) + 17}{2 \cdot 2} : \frac{7 + 17}{4}

q = \frac{- ( - 7 ) - 17}{2 \cdot 2} : \frac{7 - 17}{4}

The solutions to the quadratic equation are:

q = \frac{7 + 17}{4}, q = \frac{7 - 17}{4}

x^{2} + 12 x - 70 = 9 x

2 c^{2} + 3 c - 1 = 0

λ^{2} - 5 λ + 6 (1) = 0

x^{2} - 2 x = 2 x + 12

x^{2} - 11 x + 15 = 0

What are the solutions to the equation 2q^2-4q+4=3q ?
The solutions to the equation 2q^2-4q+4=3q are q=(7+sqrt(17))/4 ,q=(7-sqrt(17))/4
Find the zeros of 2q^2-4q+4=3q
The zeros of 2q^2-4q+4=3q are q=(7+sqrt(17))/4 ,q=(7-sqrt(17))/4