What are the solutions to the equation -5q^2+9q+13=-6q^2 ?

The solutions to the equation -5q^2+9q+13=-6q^2 are q=(-9+sqrt(29))/2 ,q=(-9-sqrt(29))/2

Find the zeros of -5q^2+9q+13=-6q^2

The zeros of -5q^2+9q+13=-6q^2 are q=(-9+sqrt(29))/2 ,q=(-9-sqrt(29))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

- 5 q^{2} + 9 q + 13 = - 6 q^{2}

q = \frac{- 9 + 29}{2}, q = \frac{- 9 - 29}{2}

Decimal Notation

q = - 1.80741 \dots, q = - 7.19258 \dots

Solution steps

- 5 q^{2} + 9 q + 13 = - 6 q^{2}

Move

6 q^{2}

to the left side

q^{2} + 9 q + 13 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

9^{2} - 4 \cdot 1 \cdot 13 = 29

q_{1, 2} = \frac{- 9 \pm 29}{2 \cdot 1}

Separate the solutions

q_{1} = \frac{- 9 + 29}{2 \cdot 1}, q_{2} = \frac{- 9 - 29}{2 \cdot 1}

q = \frac{- 9 + 29}{2 \cdot 1} : \frac{- 9 + 29}{2}

q = \frac{- 9 - 29}{2 \cdot 1} : \frac{- 9 - 29}{2}

The solutions to the quadratic equation are:

q = \frac{- 9 + 29}{2}, q = \frac{- 9 - 29}{2}

y, x^{3} + x^{2} + 2 x y^{2} + 2 y^{2} - 4 x - 4 = 0

x, z = 3 x^{2} y - 5 x y^{4} + 8 x - 7 y + 9

\frac{1}{16} = N^{2}

s (s + 8) = 0

2 (x + 9)^{2} - 18 = 374

What are the solutions to the equation -5q^2+9q+13=-6q^2 ?
The solutions to the equation -5q^2+9q+13=-6q^2 are q=(-9+sqrt(29))/2 ,q=(-9-sqrt(29))/2
Find the zeros of -5q^2+9q+13=-6q^2
The zeros of -5q^2+9q+13=-6q^2 are q=(-9+sqrt(29))/2 ,q=(-9-sqrt(29))/2