What are the solutions to the equation 1/5 q^2+10=80-3q-2q^2 ?

The solutions to the equation 1/5 q^2+10=80-3q-2q^2 are q=-70/11 ,q=5

Find the zeros of 1/5 q^2+10=80-3q-2q^2

The zeros of 1/5 q^2+10=80-3q-2q^2 are q=-70/11 ,q=5

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\frac{1}{5} q^{2} + 10 = 80 - 3 q - 2 q^{2}

q = - \frac{70}{11}, q = 5

Decimal Notation

q = - 6.36363 \dots, q = 5

Solution steps

\frac{1}{5} q^{2} + 10 = 80 - 3 q - 2 q^{2}

Multiply both sides by

5

q^{2} + 50 = 400 - 15 q - 10 q^{2}

Switch sides

400 - 15 q - 10 q^{2} = q^{2} + 50

Move

50

to the left side

- 10 q^{2} - 15 q + 350 = q^{2}

Move

q^{2}

to the left side

- 11 q^{2} - 15 q + 350 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 ( - 11 ) \cdot 350}{2 ( - 11 )}

(- 15)^{2} - 4 (- 11) \cdot 350 = 125

q_{1, 2} = \frac{- ( - 15 ) \pm 125}{2 ( - 11 )}

Separate the solutions

q_{1} = \frac{- ( - 15 ) + 125}{2 ( - 11 )}, q_{2} = \frac{- ( - 15 ) - 125}{2 ( - 11 )}

q = \frac{- ( - 15 ) + 125}{2 ( - 11 )} : - \frac{70}{11}

q = \frac{- ( - 15 ) - 125}{2 ( - 11 )} : 5

The solutions to the quadratic equation are:

q = - \frac{70}{11}, q = 5

(4 - x^{2})^{2}

\frac{x}{7} > - 1

\frac{1}{i}

\frac{x}{7} = \frac{9}{2}

\frac{5}{9} y - 4 = 6

What are the solutions to the equation 1/5 q^2+10=80-3q-2q^2 ?
The solutions to the equation 1/5 q^2+10=80-3q-2q^2 are q=-70/11 ,q=5
Find the zeros of 1/5 q^2+10=80-3q-2q^2
The zeros of 1/5 q^2+10=80-3q-2q^2 are q=-70/11 ,q=5