What are the solutions to the equation 3x^2-13x-5=4x+1 ?

The solutions to the equation 3x^2-13x-5=4x+1 are x=6,x=-1/3

Find the zeros of 3x^2-13x-5=4x+1

The zeros of 3x^2-13x-5=4x+1 are x=6,x=-1/3

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3 x^{2} - 13 x - 5 = 4 x + 1

x = 6, x = - \frac{1}{3}

Decimal Notation

x = 6, x = - 0.33333 \dots

Solution steps

3 x^{2} - 13 x - 5 = 4 x + 1

Move

1

to the left side

3 x^{2} - 13 x - 6 = 4 x

Move

4 x

to the left side

3 x^{2} - 17 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 3 ( - 6 )}{2 \cdot 3}

(- 17)^{2} - 4 \cdot 3 (- 6) = 19

x_{1, 2} = \frac{- ( - 17 ) \pm 19}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 17 ) + 19}{2 \cdot 3}, x_{2} = \frac{- ( - 17 ) - 19}{2 \cdot 3}

x = \frac{- ( - 17 ) + 19}{2 \cdot 3} : 6

x = \frac{- ( - 17 ) - 19}{2 \cdot 3} : - \frac{1}{3}

The solutions to the quadratic equation are:

x = 6, x = - \frac{1}{3}

1 - 0.4 x + 0.04 x^{2} = 0

(x + 3)^{2} + 2 = 0

144 x^{2} = 25

x, y^{2} + 4 x^{2} = 0

a, a^{2} = b^{2} + c^{2} + 2 c n

What are the solutions to the equation 3x^2-13x-5=4x+1 ?
The solutions to the equation 3x^2-13x-5=4x+1 are x=6,x=-1/3
Find the zeros of 3x^2-13x-5=4x+1
The zeros of 3x^2-13x-5=4x+1 are x=6,x=-1/3