What are the solutions to the equation 4a^2+5a-26=0 ?

The solutions to the equation 4a^2+5a-26=0 are a=2,a=-13/4

Find the zeros of 4a^2+5a-26=0

The zeros of 4a^2+5a-26=0 are a=2,a=-13/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 a^{2} + 5 a - 26 = 0

a = 2, a = - \frac{13}{4}

Solution steps

4 a^{2} + 5 a - 26 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 4 ( - 26 )}{2 \cdot 4}

5^{2} - 4 \cdot 4 (- 26) = 21

a_{1, 2} = \frac{- 5 \pm 21}{2 \cdot 4}

Separate the solutions

a_{1} = \frac{- 5 + 21}{2 \cdot 4}, a_{2} = \frac{- 5 - 21}{2 \cdot 4}

a = \frac{- 5 + 21}{2 \cdot 4} : 2

a = \frac{- 5 - 21}{2 \cdot 4} : - \frac{13}{4}

The solutions to the quadratic equation are:

a = 2, a = - \frac{13}{4}

lo g_{8} (2) + lo g_{8} (4)

1420 (\frac{107}{143})^{7} + 70

x^{2} + 6 < 0

x^{3} + x^{2} - 25 x - 25

9 x^{8} - 30 x^{4} + 25

What are the solutions to the equation 4a^2+5a-26=0 ?
The solutions to the equation 4a^2+5a-26=0 are a=2,a=-13/4
Find the zeros of 4a^2+5a-26=0
The zeros of 4a^2+5a-26=0 are a=2,a=-13/4