What are the solutions to the equation 6r^2-3r-4=r^2 ?

The solutions to the equation 6r^2-3r-4=r^2 are r=(3+sqrt(89))/(10),r=(3-sqrt(89))/(10)

Find the zeros of 6r^2-3r-4=r^2

The zeros of 6r^2-3r-4=r^2 are r=(3+sqrt(89))/(10),r=(3-sqrt(89))/(10)

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6 r^{2} - 3 r - 4 = r^{2}

r = \frac{3 + 89}{10}, r = \frac{3 - 89}{10}

Decimal Notation

r = 1.24339 \dots, r = - 0.64339 \dots

Solution steps

6 r^{2} - 3 r - 4 = r^{2}

Move

r^{2}

to the left side

5 r^{2} - 3 r - 4 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 5 (- 4) = 89

r_{1, 2} = \frac{- ( - 3 ) \pm 89}{2 \cdot 5}

Separate the solutions

r_{1} = \frac{- ( - 3 ) + 89}{2 \cdot 5}, r_{2} = \frac{- ( - 3 ) - 89}{2 \cdot 5}

r = \frac{- ( - 3 ) + 89}{2 \cdot 5} : \frac{3 + 89}{10}

r = \frac{- ( - 3 ) - 89}{2 \cdot 5} : \frac{3 - 89}{10}

The solutions to the quadratic equation are:

r = \frac{3 + 89}{10}, r = \frac{3 - 89}{10}

2 x - 0.2 x^{2} = \frac{1}{2} x

2 x^{2} = 13

2 x^{2} - 4 x - 1 = 0

(x - 13)^{2} = - \frac{1}{3}

3 y - y^{2} + 1 = 0

What are the solutions to the equation 6r^2-3r-4=r^2 ?
The solutions to the equation 6r^2-3r-4=r^2 are r=(3+sqrt(89))/(10),r=(3-sqrt(89))/(10)
Find the zeros of 6r^2-3r-4=r^2
The zeros of 6r^2-3r-4=r^2 are r=(3+sqrt(89))/(10),r=(3-sqrt(89))/(10)