What are the solutions to the equation 4q^2-15=(1+4q)/(30) ?

The solutions to the equation 4q^2-15=(1+4q)/(30) are q=(1+sqrt(13531))/(60),q=(1-sqrt(13531))/(60)

Find the zeros of 4q^2-15=(1+4q)/(30)

The zeros of 4q^2-15=(1+4q)/(30) are q=(1+sqrt(13531))/(60),q=(1-sqrt(13531))/(60)

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4 q^{2} - 15 = \frac{1 + 4 q}{30}

q = \frac{1 + 13531}{60}, q = \frac{1 - 13531}{60}

Decimal Notation

q = 1.95538 \dots, q = - 1.92204 \dots

Solution steps

4 q^{2} - 15 = \frac{1 + 4 q}{30}

Multiply both sides by

30

120 q^{2} - 450 = 1 + 4 q

Move

4 q

to the left side

120 q^{2} - 450 - 4 q = 1

Move

1

to the left side

120 q^{2} - 4 q - 451 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 120 ( - 451 )}{2 \cdot 120}

(- 4)^{2} - 4 \cdot 120 (- 451) = 413531

q_{1, 2} = \frac{- ( - 4 ) \pm 4 13531}{2 \cdot 120}

Separate the solutions

q_{1} = \frac{- ( - 4 ) + 4 13531}{2 \cdot 120}, q_{2} = \frac{- ( - 4 ) - 4 13531}{2 \cdot 120}

q = \frac{- ( - 4 ) + 4 13531}{2 \cdot 120} : \frac{1 + 13531}{60}

q = \frac{- ( - 4 ) - 4 13531}{2 \cdot 120} : \frac{1 - 13531}{60}

The solutions to the quadratic equation are:

q = \frac{1 + 13531}{60}, q = \frac{1 - 13531}{60}

8 x^{2} - 6 x = - 1

3 x = 10 - x^{2}

30 x - 0.1 x^{2} = 2250

(- 1 - x) (- x) + 6 = 0

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

What are the solutions to the equation 4q^2-15=(1+4q)/(30) ?
The solutions to the equation 4q^2-15=(1+4q)/(30) are q=(1+sqrt(13531))/(60),q=(1-sqrt(13531))/(60)
Find the zeros of 4q^2-15=(1+4q)/(30)
The zeros of 4q^2-15=(1+4q)/(30) are q=(1+sqrt(13531))/(60),q=(1-sqrt(13531))/(60)