What are the solutions to the equation 6x^2-x+1=x^2+x+4 ?

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

Find the zeros of 6x^2-x+1=x^2+x+4

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

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6 x^{2} - x + 1 = x^{2} + x + 4

x = 1, x = - \frac{3}{5}

Decimal Notation

x = 1, x = - 0.6

Solution steps

6 x^{2} - x + 1 = x^{2} + x + 4

Move

4

to the left side

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

Move

x

to the left side

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

Move

x^{2}

to the left side

5 x^{2} - 2 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

(- 2)^{2} - 4 \cdot 5 (- 3) = 8

x_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 8}{2 \cdot 5}, x_{2} = \frac{- ( - 2 ) - 8}{2 \cdot 5}

x = \frac{- ( - 2 ) + 8}{2 \cdot 5} : 1

x = \frac{- ( - 2 ) - 8}{2 \cdot 5} : - \frac{3}{5}

The solutions to the quadratic equation are:

x = 1, x = - \frac{3}{5}

y, x^{2} - 6 x y + y^{2} = 6

9 x^{2} + 6 x + 4

x, z = 64 - x^{2} - y^{2}

2 x + c = 3 x^{2} - c x - 5, c

2 X^{2} + 7 X + 6 = 0

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