What are the solutions to the equation 18q^2-32q=7 ?

The solutions to the equation 18q^2-32q=7 are q=(16+sqrt(382))/(18),q=(16-sqrt(382))/(18)

Find the zeros of 18q^2-32q=7

The zeros of 18q^2-32q=7 are q=(16+sqrt(382))/(18),q=(16-sqrt(382))/(18)

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18 q^{2} - 32 q = 7

q = \frac{16 + 382}{18}, q = \frac{16 - 382}{18}

Decimal Notation

q = 1.97471 \dots, q = - 0.19693 \dots

Solution steps

18 q^{2} - 32 q = 7

Move

7

to the left side

18 q^{2} - 32 q - 7 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 18 ( - 7 )}{2 \cdot 18}

(- 32)^{2} - 4 \cdot 18 (- 7) = 2382

q_{1, 2} = \frac{- ( - 32 ) \pm 2 382}{2 \cdot 18}

Separate the solutions

q_{1} = \frac{- ( - 32 ) + 2 382}{2 \cdot 18}, q_{2} = \frac{- ( - 32 ) - 2 382}{2 \cdot 18}

q = \frac{- ( - 32 ) + 2 382}{2 \cdot 18} : \frac{16 + 382}{18}

q = \frac{- ( - 32 ) - 2 382}{2 \cdot 18} : \frac{16 - 382}{18}

The solutions to the quadratic equation are:

q = \frac{16 + 382}{18}, q = \frac{16 - 382}{18}

x^{2} - 8 x + 4 = 0

x, 9 x^{2} - 16 y^{2} - 18 x - 64 y - 55 = 0

2 (x + 3) (x + 2) + 5 = (x + 3) (x + 1) + 6

a (b - c) x^{2} + b (c - a) x + c (a - b) = 0

2 x^{2} = 4 x + 1

What are the solutions to the equation 18q^2-32q=7 ?
The solutions to the equation 18q^2-32q=7 are q=(16+sqrt(382))/(18),q=(16-sqrt(382))/(18)
Find the zeros of 18q^2-32q=7
The zeros of 18q^2-32q=7 are q=(16+sqrt(382))/(18),q=(16-sqrt(382))/(18)