What are the solutions to the equation 2r^2-6r-8=3 ?

The solutions to the equation 2r^2-6r-8=3 are r=(3+sqrt(31))/2 ,r=(3-sqrt(31))/2

Find the zeros of 2r^2-6r-8=3

The zeros of 2r^2-6r-8=3 are r=(3+sqrt(31))/2 ,r=(3-sqrt(31))/2

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2 r^{2} - 6 r - 8 = 3

r = \frac{3 + 31}{2}, r = \frac{3 - 31}{2}

Decimal Notation

r = 4.28388 \dots, r = - 1.28388 \dots

Solution steps

2 r^{2} - 6 r - 8 = 3

Move

3

to the left side

2 r^{2} - 6 r - 11 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 2 ( - 11 )}{2 \cdot 2}

(- 6)^{2} - 4 \cdot 2 (- 11) = 231

r_{1, 2} = \frac{- ( - 6 ) \pm 2 31}{2 \cdot 2}

Separate the solutions

r_{1} = \frac{- ( - 6 ) + 2 31}{2 \cdot 2}, r_{2} = \frac{- ( - 6 ) - 2 31}{2 \cdot 2}

r = \frac{- ( - 6 ) + 2 31}{2 \cdot 2} : \frac{3 + 31}{2}

r = \frac{- ( - 6 ) - 2 31}{2 \cdot 2} : \frac{3 - 31}{2}

The solutions to the quadratic equation are:

r = \frac{3 + 31}{2}, r = \frac{3 - 31}{2}

14 < - 2 a + 6 - 2 a

- 26

3 w^{2} - 17 w + 10 = 0

5 z^{2} + 9 z - 2 = 0

7 (7 x + 8) = 5 (6 x + 7) + 59

What are the solutions to the equation 2r^2-6r-8=3 ?
The solutions to the equation 2r^2-6r-8=3 are r=(3+sqrt(31))/2 ,r=(3-sqrt(31))/2
Find the zeros of 2r^2-6r-8=3
The zeros of 2r^2-6r-8=3 are r=(3+sqrt(31))/2 ,r=(3-sqrt(31))/2