What are the solutions to the equation 2a^2+3a-5=-1 ?

The solutions to the equation 2a^2+3a-5=-1 are a=(-3+sqrt(41))/4 ,a=(-3-sqrt(41))/4

Find the zeros of 2a^2+3a-5=-1

The zeros of 2a^2+3a-5=-1 are a=(-3+sqrt(41))/4 ,a=(-3-sqrt(41))/4

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2 a^{2} + 3 a - 5 = - 1

a = \frac{- 3 + 41}{4}, a = \frac{- 3 - 41}{4}

Solution steps

2 a^{2} + 3 a - 5 = - 1

Move

1

to the left side

2 a^{2} + 3 a - 4 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 2 (- 4) = 41

a_{1, 2} = \frac{- 3 \pm 41}{2 \cdot 2}

Separate the solutions

a_{1} = \frac{- 3 + 41}{2 \cdot 2}, a_{2} = \frac{- 3 - 41}{2 \cdot 2}

a = \frac{- 3 + 41}{2 \cdot 2} : \frac{- 3 + 41}{4}

a = \frac{- 3 - 41}{2 \cdot 2} : \frac{- 3 - 41}{4}

The solutions to the quadratic equation are:

a = \frac{- 3 + 41}{4}, a = \frac{- 3 - 41}{4}

x, u = x^{2} - 2

2 x (x + 3) = 5

6 x^{2} = 7 - 11 x

(y + 5)^{2} - 25 = 0

(r + 7)^{2} + 9 = 0

What are the solutions to the equation 2a^2+3a-5=-1 ?
The solutions to the equation 2a^2+3a-5=-1 are a=(-3+sqrt(41))/4 ,a=(-3-sqrt(41))/4
Find the zeros of 2a^2+3a-5=-1
The zeros of 2a^2+3a-5=-1 are a=(-3+sqrt(41))/4 ,a=(-3-sqrt(41))/4