What are the solutions to the equation 2q^2-4q=7 ?

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

Find the zeros of 2q^2-4q=7

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

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2 q^{2} - 4 q = 7

q = \frac{2 + 3 2}{2}, q = \frac{2 - 3 2}{2}

Decimal Notation

q = 3.12132 \dots, q = - 1.12132 \dots

Solution steps

2 q^{2} - 4 q = 7

Move

7

to the left side

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

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 2 (- 7) = 62

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

Separate the solutions

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

q = \frac{- ( - 4 ) + 6 2}{2 \cdot 2} : \frac{2 + 3 2}{2}

q = \frac{- ( - 4 ) - 6 2}{2 \cdot 2} : \frac{2 - 3 2}{2}

The solutions to the quadratic equation are:

q = \frac{2 + 3 2}{2}, q = \frac{2 - 3 2}{2}

n^{2} - 4 n + 29 = 0

X^{2} + 2 X - 24 = 0

- 0.0625 x^{2} + 2 x + 50 = 0

4 x (x + 5) = 6 x - 4 x^{2} - 3

6 x^{2} + 5 x - 4 = 0

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