What are the solutions to the equation 2q^2+3q-8=0 ?

The solutions to the equation 2q^2+3q-8=0 are q=(-3+sqrt(73))/4 ,q=(-3-sqrt(73))/4

Find the zeros of 2q^2+3q-8=0

The zeros of 2q^2+3q-8=0 are q=(-3+sqrt(73))/4 ,q=(-3-sqrt(73))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 q^{2} + 3 q - 8 = 0

q = \frac{- 3 + 73}{4}, q = \frac{- 3 - 73}{4}

Decimal Notation

q = 1.38600 \dots, q = - 2.88600 \dots

Solution steps

2 q^{2} + 3 q - 8 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 8 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 8) = 73

q_{1, 2} = \frac{- 3 \pm 73}{2 \cdot 2}

Separate the solutions

q_{1} = \frac{- 3 + 73}{2 \cdot 2}, q_{2} = \frac{- 3 - 73}{2 \cdot 2}

q = \frac{- 3 + 73}{2 \cdot 2} : \frac{- 3 + 73}{4}

q = \frac{- 3 - 73}{2 \cdot 2} : \frac{- 3 - 73}{4}

The solutions to the quadratic equation are:

q = \frac{- 3 + 73}{4}, q = \frac{- 3 - 73}{4}

\frac{11}{6} - \frac{5}{8}

- 2 (x + 13) + 37 \leq 5 x - 17

x (x - 2) = 15

(x - 5) (x - 2) (x + 7) > 0

\frac{1}{3} x^{2} - \frac{1}{9} x + \frac{10}{9} = 0

What are the solutions to the equation 2q^2+3q-8=0 ?
The solutions to the equation 2q^2+3q-8=0 are q=(-3+sqrt(73))/4 ,q=(-3-sqrt(73))/4
Find the zeros of 2q^2+3q-8=0
The zeros of 2q^2+3q-8=0 are q=(-3+sqrt(73))/4 ,q=(-3-sqrt(73))/4