What are the solutions to the equation 2a^2+11a+10=0 ?

The solutions to the equation 2a^2+11a+10=0 are a=(-11+sqrt(41))/4 ,a=(-11-sqrt(41))/4

Find the zeros of 2a^2+11a+10=0

The zeros of 2a^2+11a+10=0 are a=(-11+sqrt(41))/4 ,a=(-11-sqrt(41))/4

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2 a^{2} + 11 a + 10 = 0

a = \frac{- 11 + 41}{4}, a = \frac{- 11 - 41}{4}

Decimal Notation

a = - 1.14921 \dots, a = - 4.35078 \dots

Solution steps

2 a^{2} + 11 a + 10 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 2 \cdot 10}{2 \cdot 2}

1 1^{2} - 4 \cdot 2 \cdot 10 = 41

a_{1, 2} = \frac{- 11 \pm 41}{2 \cdot 2}

Separate the solutions

a_{1} = \frac{- 11 + 41}{2 \cdot 2}, a_{2} = \frac{- 11 - 41}{2 \cdot 2}

a = \frac{- 11 + 41}{2 \cdot 2} : \frac{- 11 + 41}{4}

a = \frac{- 11 - 41}{2 \cdot 2} : \frac{- 11 - 41}{4}

The solutions to the quadratic equation are:

a = \frac{- 11 + 41}{4}, a = \frac{- 11 - 41}{4}

3 x^{2} + 64 = \frac{2 x ^{2}}{2}

3 w^{2} + 12 w - 15 = 0

3 x^{2} + 4 x - 12 = 13

x^{2} - 5 x = 21

h, (2 - t) ((5 - h - t) (5 + h - t)) + h^{2} = 0

What are the solutions to the equation 2a^2+11a+10=0 ?
The solutions to the equation 2a^2+11a+10=0 are a=(-11+sqrt(41))/4 ,a=(-11-sqrt(41))/4
Find the zeros of 2a^2+11a+10=0
The zeros of 2a^2+11a+10=0 are a=(-11+sqrt(41))/4 ,a=(-11-sqrt(41))/4