What are the solutions to the equation 12l^2+48l-826=0 ?

The solutions to the equation 12l^2+48l-826=0 are l=(-12+sqrt(2622))/6 ,l=-(12+sqrt(2622))/6

Find the zeros of 12l^2+48l-826=0

The zeros of 12l^2+48l-826=0 are l=(-12+sqrt(2622))/6 ,l=-(12+sqrt(2622))/6

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12 l^{2} + 48 l - 826 = 0

l = \frac{- 12 + 2622}{6}, l = - \frac{12 + 2622}{6}

Decimal Notation

l = 6.53424 \dots, l = - 10.53424 \dots

Solution steps

12 l^{2} + 48 l - 826 = 0

Solve with the quadratic formula

l_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 \cdot 12 ( - 826 )}{2 \cdot 12}

4 8^{2} - 4 \cdot 12 (- 826) = 42622

l_{1, 2} = \frac{- 48 \pm 4 2622}{2 \cdot 12}

Separate the solutions

l_{1} = \frac{- 48 + 4 2622}{2 \cdot 12}, l_{2} = \frac{- 48 - 4 2622}{2 \cdot 12}

l = \frac{- 48 + 4 2622}{2 \cdot 12} : \frac{- 12 + 2622}{6}

l = \frac{- 48 - 4 2622}{2 \cdot 12} : - \frac{12 + 2622}{6}

The solutions to the quadratic equation are:

l = \frac{- 12 + 2622}{6}, l = - \frac{12 + 2622}{6}

441

x^{2} + 11 x + 2 = 0

x^{- \frac{5}{3}}

\frac{5}{12} (x + 2) = \frac{2}{3} (2 x + 1) + \frac{1}{6}

\frac{81 x ^{2} y ^{3}}{9 x y ^{2}}

What are the solutions to the equation 12l^2+48l-826=0 ?
The solutions to the equation 12l^2+48l-826=0 are l=(-12+sqrt(2622))/6 ,l=-(12+sqrt(2622))/6
Find the zeros of 12l^2+48l-826=0
The zeros of 12l^2+48l-826=0 are l=(-12+sqrt(2622))/6 ,l=-(12+sqrt(2622))/6