What are the solutions to the equation 12g^2-31g=10 ?

The solutions to the equation 12g^2-31g=10 are g=(31+sqrt(1441))/(24),g=(31-sqrt(1441))/(24)

Find the zeros of 12g^2-31g=10

The zeros of 12g^2-31g=10 are g=(31+sqrt(1441))/(24),g=(31-sqrt(1441))/(24)

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12 g^{2} - 31 g = 10

g = \frac{31 + 1441}{24}, g = \frac{31 - 1441}{24}

Decimal Notation

g = 2.87335 \dots, g = - 0.29002 \dots

Solution steps

12 g^{2} - 31 g = 10

Move

10

to the left side

12 g^{2} - 31 g - 10 = 0

Solve with the quadratic formula

g_{1, 2} = \frac{- ( - 31 ) \pm ( - 31 ) ^{2} - 4 \cdot 12 ( - 10 )}{2 \cdot 12}

(- 31)^{2} - 4 \cdot 12 (- 10) = 1441

g_{1, 2} = \frac{- ( - 31 ) \pm 1441}{2 \cdot 12}

Separate the solutions

g_{1} = \frac{- ( - 31 ) + 1441}{2 \cdot 12}, g_{2} = \frac{- ( - 31 ) - 1441}{2 \cdot 12}

g = \frac{- ( - 31 ) + 1441}{2 \cdot 12} : \frac{31 + 1441}{24}

g = \frac{- ( - 31 ) - 1441}{2 \cdot 12} : \frac{31 - 1441}{24}

The solutions to the quadratic equation are:

g = \frac{31 + 1441}{24}, g = \frac{31 - 1441}{24}

lo g_{6} (4) + lo g_{6} (3) - lo g_{6} (2) = x

3 y^{2} - 4 y - 2 = 0

2 - 2 i

9 < 3 (- \frac{1}{3} p + 6)

What are the solutions to the equation 12g^2-31g=10 ?
The solutions to the equation 12g^2-31g=10 are g=(31+sqrt(1441))/(24),g=(31-sqrt(1441))/(24)
Find the zeros of 12g^2-31g=10
The zeros of 12g^2-31g=10 are g=(31+sqrt(1441))/(24),g=(31-sqrt(1441))/(24)