What are the solutions to the equation 11a^2+12a=15 ?

The solutions to the equation 11a^2+12a=15 are a=(-6+sqrt(201))/(11),a=-(6+sqrt(201))/(11)

Find the zeros of 11a^2+12a=15

The zeros of 11a^2+12a=15 are a=(-6+sqrt(201))/(11),a=-(6+sqrt(201))/(11)

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11 a^{2} + 12 a = 15

a = \frac{- 6 + 201}{11}, a = - \frac{6 + 201}{11}

Decimal Notation

a = 0.74340 \dots, a = - 1.83431 \dots

Solution steps

11 a^{2} + 12 a = 15

Move

15

to the left side

11 a^{2} + 12 a - 15 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 11 ( - 15 )}{2 \cdot 11}

1 2^{2} - 4 \cdot 11 (- 15) = 2201

a_{1, 2} = \frac{- 12 \pm 2 201}{2 \cdot 11}

Separate the solutions

a_{1} = \frac{- 12 + 2 201}{2 \cdot 11}, a_{2} = \frac{- 12 - 2 201}{2 \cdot 11}

a = \frac{- 12 + 2 201}{2 \cdot 11} : \frac{- 6 + 201}{11}

a = \frac{- 12 - 2 201}{2 \cdot 11} : - \frac{6 + 201}{11}

The solutions to the quadratic equation are:

a = \frac{- 6 + 201}{11}, a = - \frac{6 + 201}{11}

- x^{2} - 8 = 0

0.004 y^{2} - 0.01 y + 121 = 50

60 + \frac{1}{2} ((a - 200) (5 - \frac{a}{40})) = 0

\frac{1}{4} a^{2 n} = x^{2}

(3 x + 7)^{2} = 22

What are the solutions to the equation 11a^2+12a=15 ?
The solutions to the equation 11a^2+12a=15 are a=(-6+sqrt(201))/(11),a=-(6+sqrt(201))/(11)
Find the zeros of 11a^2+12a=15
The zeros of 11a^2+12a=15 are a=(-6+sqrt(201))/(11),a=-(6+sqrt(201))/(11)