What are the solutions to the equation log_{10}(12)(x^2-x)=1 ?

The solutions to the equation log_{10}(12)(x^2-x)=1 are

Find the zeros of log_{10}(12)(x^2-x)=1

The zeros of log_{10}(12)(x^2-x)=1 are

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lo g_{10} (12) (x^{2} - x) = 1

x = \frac{lo g _{10} ( 12 ) + lo g _{10} ( 12 ) ^{2} + 4 lo g _{10} ( 12 )}{2 lo g _{10} ( 12 )}, x = \frac{lo g _{10} ( 12 ) - lo g _{10} ( 12 ) ^{2} + 4 lo g _{10} ( 12 )}{2 lo g _{10} ( 12 )}

Decimal Notation

x = 1.58472 \dots, x = - 0.58472 \dots

Solution steps

lo g_{10} (12) (x^{2} - x) = 1

Expand

lo g_{10} (12) (x^{2} - x) : lo g_{10} (12) x^{2} - lo g_{10} (12) x

lo g_{10} (12) x^{2} - lo g_{10} (12) x = 1

Move

1

to the left side

lo g_{10} (12) x^{2} - lo g_{10} (12) x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - lo g _{10} ( 12 ) ) \pm ( - lo g _{10} ( 12 ) ) ^{2} - 4 lo g _{10} ( 12 ) ( - 1 )}{2 lo g _{10} ( 12 )}

(- lo g_{10} (12))^{2} - 4 lo g_{10} (12) (- 1) = lo g_{10} (12)^{2} + 4 lo g_{10} (12)

x_{1, 2} = \frac{- ( - lo g _{10} ( 12 ) ) \pm lo g _{10} ( 12 ) ^{2} + 4 lo g _{10} ( 12 )}{2 lo g _{10} ( 12 )}

Separate the solutions

x_{1} = \frac{- ( - lo g _{10} ( 12 ) ) + lo g _{10} ( 12 ) ^{2} + 4 lo g _{10} ( 12 )}{2 lo g _{10} ( 12 )}, x_{2} = \frac{- ( - lo g _{10} ( 12 ) ) - lo g _{10} ( 12 ) ^{2} + 4 lo g _{10} ( 12 )}{2 lo g _{10} ( 12 )}

x = \frac{- ( - lo g _{10} ( 12 ) ) + lo g _{10} ( 12 ) ^{2} + 4 lo g _{10} ( 12 )}{2 lo g _{10} ( 12 )} : \frac{lo g _{10} ( 12 ) + lo g _{10} ( 12 ) ^{2} + 4 lo g _{10} ( 12 )}{2 lo g _{10} ( 12 )}

x = \frac{- ( - lo g _{10} ( 12 ) ) - lo g _{10} ( 12 ) ^{2} + 4 lo g _{10} ( 12 )}{2 lo g _{10} ( 12 )} : \frac{lo g _{10} ( 12 ) - lo g _{10} ( 12 ) ^{2} + 4 lo g _{10} ( 12 )}{2 lo g _{10} ( 12 )}

The solutions to the quadratic equation are:

x = \frac{lo g _{10} ( 12 ) + lo g _{10} ( 12 ) ^{2} + 4 lo g _{10} ( 12 )}{2 lo g _{10} ( 12 )}, x = \frac{lo g _{10} ( 12 ) - lo g _{10} ( 12 ) ^{2} + 4 lo g _{10} ( 12 )}{2 lo g _{10} ( 12 )}

x^{2} - 5 x + 6 = 2 - 3

x^{2} - (1 + 2) \cdot x + 2 = 0

\frac{( x ^{2} )}{6} - \frac{( 3 x )}{6} = 3 (x - 5)

4 x^{2} - 4 a^{2} \cdot x + (a^{4} - b^{4}) = 0

x, x^{2} - 10 x + y^{2} - 20 y = - 125

What are the solutions to the equation log_{10}(12)(x^2-x)=1 ?
The solutions to the equation log_{10}(12)(x^2-x)=1 are
Find the zeros of log_{10}(12)(x^2-x)=1
The zeros of log_{10}(12)(x^2-x)=1 are