What are the solutions to the equation 2x^2+7x-13=-10 ?

The solutions to the equation 2x^2+7x-13=-10 are x=(-7+sqrt(73))/4 ,x=(-7-sqrt(73))/4

Find the zeros of 2x^2+7x-13=-10

The zeros of 2x^2+7x-13=-10 are x=(-7+sqrt(73))/4 ,x=(-7-sqrt(73))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{2} + 7 x - 13 = - 10

x = \frac{- 7 + 73}{4}, x = \frac{- 7 - 73}{4}

Decimal Notation

x = 0.38600 \dots, x = - 3.88600 \dots

Solution steps

2 x^{2} + 7 x - 13 = - 10

Move

10

to the left side

2 x^{2} + 7 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

7^{2} - 4 \cdot 2 (- 3) = 73

x_{1, 2} = \frac{- 7 \pm 73}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 7 + 73}{2 \cdot 2}, x_{2} = \frac{- 7 - 73}{2 \cdot 2}

x = \frac{- 7 + 73}{2 \cdot 2} : \frac{- 7 + 73}{4}

x = \frac{- 7 - 73}{2 \cdot 2} : \frac{- 7 - 73}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 73}{4}, x = \frac{- 7 - 73}{4}

24 = 4 + 37 t - 16 t^{2}

x, 4 x^{2} y + 2 y^{3} = 1 + 6 x y^{2}

k^{2} + k = 0

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

27 = x^{2}

What are the solutions to the equation 2x^2+7x-13=-10 ?
The solutions to the equation 2x^2+7x-13=-10 are x=(-7+sqrt(73))/4 ,x=(-7-sqrt(73))/4
Find the zeros of 2x^2+7x-13=-10
The zeros of 2x^2+7x-13=-10 are x=(-7+sqrt(73))/4 ,x=(-7-sqrt(73))/4