What are the solutions to the equation 2q^2+6q-7=0 ?

The solutions to the equation 2q^2+6q-7=0 are q=(-3+sqrt(23))/2 ,q=-(3+sqrt(23))/2

Find the zeros of 2q^2+6q-7=0

The zeros of 2q^2+6q-7=0 are q=(-3+sqrt(23))/2 ,q=-(3+sqrt(23))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 q^{2} + 6 q - 7 = 0

q = \frac{- 3 + 23}{2}, q = - \frac{3 + 23}{2}

Decimal Notation

q = 0.89791 \dots, q = - 3.89791 \dots

Solution steps

2 q^{2} + 6 q - 7 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 ( - 7 )}{2 \cdot 2}

6^{2} - 4 \cdot 2 (- 7) = 223

q_{1, 2} = \frac{- 6 \pm 2 23}{2 \cdot 2}

Separate the solutions

q_{1} = \frac{- 6 + 2 23}{2 \cdot 2}, q_{2} = \frac{- 6 - 2 23}{2 \cdot 2}

q = \frac{- 6 + 2 23}{2 \cdot 2} : \frac{- 3 + 23}{2}

q = \frac{- 6 - 2 23}{2 \cdot 2} : - \frac{3 + 23}{2}

The solutions to the quadratic equation are:

q = \frac{- 3 + 23}{2}, q = - \frac{3 + 23}{2}

x, f = x^{2} f + g

2 (4 x + 3)^{2} - 46 = - 12

y, x^{2} - 2 x y + 3 y^{2} = 4

x^{2} - 15 = 1

x^{2} - 2 π x - 3 π^{2} = 0

What are the solutions to the equation 2q^2+6q-7=0 ?
The solutions to the equation 2q^2+6q-7=0 are q=(-3+sqrt(23))/2 ,q=-(3+sqrt(23))/2
Find the zeros of 2q^2+6q-7=0
The zeros of 2q^2+6q-7=0 are q=(-3+sqrt(23))/2 ,q=-(3+sqrt(23))/2